Electricity-Exercise

Q1.A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is (a) 1/25 (b) 1/5 (c) 5 (d) 25

Answer

Let the resistance of the original wire be \( R\). When the wire is cut into five equal parts, the resistance of each part, denoted as \(R_n\), becomes one-fifth of the original resistance because resistance is directly proportional to the length of the wire. Hence,

Since these five parts are connected in parallel, the equivalent resistance $$ R' $$ can be found using the formula for parallel resistors. The reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances:

Substituting $$ R_n = \frac{R}{5} $$ into the expression gives:

$$ \begin{aligned} \frac{1}{R'}& = \frac{5}{\frac{R}{5}} \\\\&= \frac{5 \times 5}{R} \\\\&= \frac{25}{R} \end{aligned}$$

Taking the reciprocal, we find:

$$ R' = \frac{R}{25} $$

Therefore, the ratio of the original resistance to the equivalent resistance is:

$$\begin{aligned} \frac{R}{R'} &= \frac{R}{\frac{R}{25}} \\\\&= 25 \end{aligned}$$

Hence, the correct answer is (d) 25.

Q2. Which of the following terms does not represent electrical power in a circuit? (a) I2R (b) IR2 (c) VI (d) V2/R

Answer

The electrical power dissipated or used in a circuit is given by the formula $$ P = VI $$ where \(V \) is the voltage across the circuit and \(I \) is the current flowing through it.

Using Ohm's law, which states $$ V = IR $$ the power formula can be expressed in different equivalent forms. Substituting $$ V = IR $$ into $$ P = VI $$ gives:

$$ P = I \times (IR) = I^2 R $$

Similarly, substituting $$ I = \frac{V}{R} $$ into $$ P = VI $$ results in:

$$ P = V \times \frac{V}{R} = \frac{V^2}{R} $$

Among the options provided, \(I^2 R\), \(VI\), and \(\frac{V^2}{R}\) are all valid expressions representing electrical power. However, the term \(IR^2\) does not correspond to electrical power in a circuit because it does not emerge from the relationship among voltage, current, and resistance. Therefore, the term that does not represent electrical power is \(IR^2\), which corresponds to option (b).

Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be (a) 100 W (b) 75 W (c) 50 W (d) 25 W

Answer

The electric bulb is rated at 220 V and 100 W, which means at 220 V the power consumed is 100 W. We can first find the resistance $$ R $$ of the bulb using the formula for power in terms of voltage and resistance:

$$ P = \frac{V^2}{R} $$

Rearranging for \(R\), we get:

$$ R = \frac{V^2}{P} $$

Substituting the rated voltage and power values:

$$\begin{aligned} R &= \frac{(220)^2}{100} \\\\&= \frac{48400}{100} \\\\&= 484 \, \Omega \end{aligned}$$

Now, when the bulb is operated at 110 V, the power consumed $$ P' $$ can be calculated using the same formula:

$$\begin{aligned} P' &= \frac{V'^2}{R} \\\\&= \frac{(110)^2}{484} \\\\&= \frac{12100}{484} \\\\&= 25 \, W \end{aligned}$$

Thus, when the bulb is operated at 110 V, the power consumed will be 25 W, which matches option (d).

Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

Answer

Let's carefully analyze the problem step-by-step. Two identical wires each have resistance \(R\). First, they are connected in series, then in parallel, across the same potential difference \(V\).

The equivalent resistance in series is:

$$ R_s = R + R = 2R $$

The current through the series combination is:

$$ I_s = \frac{V}{R_s} = \frac{V}{2R} $$

The heat produced in the series combination over time $$ t $$ (taking $$ t = 1 $$ for simplicity) is given by Joule's law:

$$\begin{aligned} H_s &= I_s^2 R_s t \\\\&= \left(\frac{V}{2R}\right)^2 \times 2R \times 1 \\\\&= \frac{V^2}{4R^2} \times 2R\\\\& = \frac{V^2}{2R} \end{aligned}$$

For the parallel combination, the equivalent resistance is:

$$\begin{aligned} \frac{1}{R_p} &= \frac{1}{R} + \frac{1}{R} \\\\&= \frac{2}{R} \\\\\implies R_p &= \frac{R}{2} \end{aligned}$$

The current in the parallel combination is:

$$\begin{aligned} I_p &= \frac{V}{R_p} \\\\&= \frac{V}{R/2} \\\\&= \frac{2V}{R} \end{aligned}$$

The heat produced in the parallel combination over time $$ t = 1 $$ is:

$$\begin{aligned} H_p &= I_p^2 R_p t \\\\&= \left(\frac{2V}{R}\right)^2 \times \frac{R}{2} \times 1 \\\\&= \frac{4V^2}{R^2} \times \frac{R}{2} \\\\&= \frac{4V^2}{2R} \\\\&= \frac{2V^2}{R} \end{aligned}$$

Finally, the ratio of heat produced in series to that in parallel is:

$$\begin{aligned} \frac{H_s}{H_p} &= \frac{\frac{V^2}{2R}}{\frac{2V^2}{R}} \\\\&= \frac{V^2}{2R} \times \frac{R}{2V^2} \\\\&= \frac{1}{4} \end{aligned}$$

Thus, the heat produced ratio $$ H_s : H_p = 1 : 4 $$ The correct answer is option (c) 1:4.

Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer

A voltmeter is connected in parallel across the two points between which the potential difference is to be measured. This parallel connection is essential because the voltmeter must measure the voltage drop directly across the component without significantly altering the current flowing through the circuit. Since a voltmeter has a very high resistance, it draws a negligible current, ensuring that it does not affect the circuit operation.

Connecting the voltmeter in series would cause a large resistance to come in series with the circuit, reducing the current drastically and potentially causing incorrect readings or stopping the circuit from functioning properly. Therefore, the correct way to measure potential difference is by connecting the voltmeter in parallel across the points of interest.

Q6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer

The resistance $$ R $$ of a wire is given by the formula:

$$ R = \rho \frac{l}{A} $$

where,
\(\rho\) : is the resistivity of the material, and
\(l\) : is the length of the wire,
\(A\) : is the cross-sectional area.

The diameter of the copper wire is given as 0.5 mm, so the radius \(r\) is half of that,

$$\begin{aligned} r &= \frac{0.5}{2} \text{ mm} \\\\&= 0.25 \times 10^{-3} \text{ m} \end{aligned}$$

The cross-sectional area \(A\) for a wire of radius \(r\) is:

$$\begin{aligned} A &= \pi r^2 \\&= \pi \times (0.25 \times 10^{-3})^2 \\&= \pi \times 6.25 \times 10^{-8} \\&= 1.9635 \times 10^{-7} \, \text{m}^2 \end{aligned}$$

Rearranging the resistance formula to solve for length \(l\) :

$$ l = \frac{R \times A}{\rho} $$

Substituting the values $$ R = 10\, \Omega $$ $$ A = 1.9635 \times 10^{-7} \, \text{m}^2 $$ and $$ \rho = 1.6 \times 10^{-8}\, \Omega \cdot \text{m} $$ gives:

$$\begin{aligned} l &= \frac{10 \times 1.9635 \times 10^{-7}}{1.6 \times 10^{-8}} \\&= 122.7 \, \text{m} \end{aligned}$$

If the diameter is doubled, the new diameter becomes 1.0 mm, so the new radius is $$ 0.5 \times 10^{-3} \, \text{m} $$ The new cross-sectional area \(A'\) is:

$$\begin{aligned} A' &= \pi \times (0.5 \times 10^{-3})^2 \\&= \pi \times 25 \times 10^{-8} \\&= 7.854 \times 10^{-7} \, \text{m}^2 \end{aligned}$$

Since resistance is inversely proportional to the cross-sectional area for the same length, the new resistance \(R'\) is:

$$\begin{aligned} R' &= \rho \frac{l}{A'} \\\\&= \frac{R A}{A'} \\\\&= \frac{10 \times 1.9635 \times 10^{-7}}{7.854 \times 10^{-7}} \\\\&= 2.5\, \Omega \end{aligned}$$

Therefore, doubling the diameter decreases the resistance to one-fourth of its original value, reducing it from 10 Ω to 2.5 Ω.

Q8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer

To determine the resistance of the unknown resistor, begin by noting the given values: the battery provides a potential difference of $$ 12\, \text{V} $$ and the current in the circuit is $$ 2.5\, \text{mA} $$ To apply Ohm’s Law, convert the current to amperes: $$ 2.5\, \text{mA} = 2.5 \times 10^{-3}\, \text{A} $$

Ohm’s Law relates resistance (\(R\)), voltage (\(V\)), and current (\(I\)) as $$ R = \frac{V}{I} $$ Substituting the values, we have:

$$\begin{aligned} R &= \frac{12}{2.5 \times 10^{-3}} \\\\&= \frac{12}{0.0025} \\\\&= 4800\, \Omega \\\\&= 4.8\, \text{k}\Omega \end{aligned}$$

Thus, the value of resistance of the resistor is $$ 4.8\, \text{k}\Omega $$

Q9.A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer

The problem states that a 12 V battery is connected across an unknown resistor and a current of 2.5 mA flows through the circuit. Using Ohm's Law, the resistance \(R\) of the resistor can be calculated by the formula:

$$ R = \frac{V}{I} $$

Here, the voltage $$ V = 12 \, \text{V} $$ and current $$\begin{aligned} I &= 2.5 \, \text{mA} \\\\&= 2.5 \times 10^{-3} \, \text{A} \end{aligned}$$ Substituting these values:

$$\begin{aligned} R &= \frac{12}{2.5 \times 10^{-3}} \\\\&= \frac{12}{0.0025} \\\\&= 4800 \, \Omega \\\\&= 4.8 \, k\Omega \end{aligned}$$

Therefore, the resistance of the resistor is $$ 4.8 \, k\Omega $$

Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer

The question asks how many 176 Ω resistors connected in parallel are required to carry a current of 5 A on a 220 V line. To find this, first calculate the total equivalent resistance \(R_\text{eq}\) of the parallel combination using Ohm's law:

$$\begin{aligned} R_\text{eq} &= \frac{V}{I} \\\\&= \frac{220}{5} \\\\&= 44 \, \Omega \end{aligned}$$

Since the resistors are identical and connected in parallel, the equivalent resistance \(R_\text{eq}\) is related to the number of resistors \(n\) and the resistance of each resistor \(R = 176 \, \Omega\) by:

$$\begin{aligned} \frac{1}{R_\text{eq}} &= n \times \frac{1}{R} \\\implies n &= \frac{R}{R_\text{eq}} \\\\&= \frac{176}{44} \\\\&= 4 \end{aligned}$$

Therefore, 4 resistors of 176 Ω each connected in parallel are required to carry 5 A current at 220 V.

Q11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer

To obtain a combined resistance of 9 Ω using three resistors each of 6 Ω, connect two resistors in parallel and then connect the third resistor in series with this combination. First, find the equivalent resistance of two 6 Ω resistors in parallel:

$$\begin{aligned} \frac{1}{R_p} &= \frac{1}{6} + \frac{1}{6} \\\\&= \frac{2}{6} \\\\\implies R_p &= 3\, \Omega \end{aligned}$$

Now, add the third resistor in series:

$$\begin{aligned} R_\text{total} &= R_p + 6 \\&= 3 + 6 \\&= 9\, \Omega \end{aligned}$$

For a resistance of 4 Ω using three 6 Ω resistors, no combination of series and parallel connections can yield this exact resistance. This is because parallel combinations will result in resistances less than the smallest resistor (6 Ω), but using three resistors in parallel gives:

$$\begin{aligned} \frac{1}{R_p} &= \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\\\&= \frac{3}{6} \\\\&= \frac{1}{2} \\\\\implies R_p &= 2\, \Omega \end{aligned}$$

which is less than 4 Ω, and any series combination will increase the total resistance beyond 6 Ω. Thus, 4 Ω is not achievable with these three resistors.

Q12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Answer

The electric bulbs are rated 10 W each and intended to be used on a 220 V supply line. The maximum allowable current in the circuit is 5 A. First, calculate the total power that can be supplied without exceeding this current limit using the formula:

$$\begin{aligned} P_\text{max} &= V \times I \\&= 220 \times 5 \\&= 1100 \, \text{W} \end{aligned}$$

Since each bulb consumes 10 W, the maximum number of such bulbs that can be connected in parallel across the line without exceeding the allowable current is:

$$\begin{aligned} n &= \frac{P_\text{max}}{P_\text{bulb}} \\\\&= \frac{1100}{10} \\\\&= 110 \end{aligned}$$

Therefore, up to 110 bulbs can be connected in parallel across the 220 V line without exceeding the 5 A current limit.

Q13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer

The given hot plate is connected to a 220 V supply line and has two resistance coils, A and B, each with resistance 24 Ω. We will calculate the current in three cases: when the coils are used separately, in series, and in parallel.

When either coil is used separately, the current through the coil is given by Ohm's Law:

$$\begin{aligned} I &= \frac{V}{R} \\\\&= \frac{220}{24} \\\\&\approx 9.17 \, \text{A} \end{aligned}$$

When the two coils are connected in series, the equivalent resistance is the sum of the two resistances:

$$\begin{aligned} R_{\text{series}} &= 24 + 24 \\&= 48 \, \Omega \end{aligned}$$

The current in the series combination is:

$$\begin{aligned} I &= \frac{V}{R_{\text{series}}} \\\\&= \frac{220}{48} \\\\&\approx 4.58 \, \text{A} \end{aligned}$$

When the coils are connected in parallel, the equivalent resistance is calculated by:

$$\begin{aligned} \frac{1}{R_{\text{parallel}}} &= \frac{1}{24} + \frac{1}{24} \\\\&= \frac{2}{24} \\\\&= \frac{1}{12} \\\\\implies R_{\text{parallel}} &= 12 \, \Omega \end{aligned}$$

The current in the parallel combination is then:

$$\begin{aligned} I &= \frac{V}{R_{\text{parallel}}} \\\\&= \frac{220}{12} \\\\&\approx 18.33 \, \text{A} \end{aligned}$$

Q14.Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer

The question asks to compare the power used in a 2 Ω resistor under two different circuit conditions: (i) when connected in series with a 6 V battery and a 1 Ω resistor, and (ii) when connected in parallel with a 4 V battery and a 12 Ω resistor.

For the first case (series circuit with 6 V battery, 1 Ω and 2 Ω resistors):

The total resistance in series is:

$$\begin{aligned} R_\text{total} &= 1 + 2 \\&= 3 \, \Omega\end{aligned} $$

The current flowing through the circuit is:

$$\begin{aligned} I &= \frac{V}{R_\text{total}} \\\\&= \frac{6}{3} \\\\&= 2 \, \text{A}\end{aligned} $$

The power used in the 2 Ω resistor is then:

$$\begin{aligned} P &= I^2 R \\&= (2)^2 \times 2 \\&= 4 \times 2 \\&= 8 \, \text{W}\end{aligned} $$

For the second case (parallel circuit with 4 V battery, 12 Ω and 2 Ω resistors):

The voltage across each resistor in parallel is the same as that of the battery, 4 V. Therefore, power used in the 2 Ω resistor is:

$$\begin{aligned} P &= \frac{V^2}{R} \\\\&= \frac{4^2}{2} \\\\&= \frac{16}{2} \\\\&= 8 \, \text{W}\end{aligned} $$

Thus, the power used in the 2 Ω resistor is the same in both cases, equal to 8 watts.

Q15.Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer

Two lamps rated 100 W and 60 W at 220 V are connected in parallel to a 220 V supply. The total power consumed by both lamps is the sum of their individual powers:

$$\begin{aligned} P_\text{total} &= 100 + 60 \\&= 160 \, W\end{aligned} $$

The current drawn from the supply line can be calculated using the power formula $$ P = VI $$ where,
\(V = 220 \, V\) is the supply voltage and
\(I\) is the current:

$$\begin{aligned} I &= \frac{P_\text{total}}{V} \\\\&= \frac{160}{220} \\\\&\approx 0.73 \, A\end{aligned} $$

Therefore, the total current drawn from the line by the two lamps connected in parallel is approximately 0.73 A.

Q16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer

The energy consumed by an electrical device is calculated using the formula:

$$ E = P \times t $$

where,
\(E\) is the energy in watt-hours (Wh),
\(P\) is the power rating in watts (W), and
\(t\) is the time in hours (h).

For the 250 W TV operating for 1 hour, the energy consumed is:

$$\begin{aligned} E_\text{TV} &= 250 \times 1 \\&= 250 \, \text{Wh}\end{aligned} $$

For the 1200 W toaster operating for 10 minutes (which is \(\frac{10}{60} = \frac{1}{6}\) hours), the energy consumed is:

$$\begin{aligned} E_\text{toaster} &= 1200 \times \frac{10}{60} \\\\&= 1200 \times \frac{1}{6} \\\\&= 200 \, \text{Wh}\end{aligned} $$

Comparing the two, the TV consumes more energy (250 Wh) than the toaster (200 Wh) over their respective operating times.

Q17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Answer

The problem involves calculating the rate at which heat is developed in an electric heater with resistance 44 Ω, drawing a current of 5 A for 2 hours. The heat generated per second (power) in an electrical resistor is given by the formula:

$$ H = I^2 R $$

Substituting the given values, the heat generated per second (power) is:

$$\begin{aligned} H &= (5)^2 \times 44 \\&= 25 \times 44 \\&= 1100 \, \text{W}\end{aligned} $$

This indicates that the heater develops heat at a rate of 1100 watts, meaning it produces 1100 joules of heat energy every second while operating.

Q18. Explain the following. (a) Why is the tungsten used almost exclusively for filament of electric lamps? (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer

(a) Why is tungsten used almost exclusively for filaments of electric lamps?

Tungsten is used for lamp filaments because it has an exceptionally high melting point, around 3422°C, allowing it to withstand the intense heat generated when electric current passes through it without melting. Additionally, tungsten has a low evaporation rate (low vapor pressure), so the filament lasts longer without thinning quickly. Its high tensile strength and ductility enable it to be drawn into fine wires that are strong enough to maintain their shape even at high temperatures. These properties make tungsten the ideal material for filaments that need to glow brightly for extended periods without failing.

(b) Why are conductors of electric heating devices made of an alloy rather than a pure metal?

Conductors in heating devices like toasters and electric irons are made from alloys because alloys have higher resistances and better mechanical strength at high temperatures compared to pure metals. Alloys also resist oxidation and deformation when heated repeatedly, which improves durability and safety. Moreover, the resistance of alloys does not change drastically with temperature, allowing these devices to generate heat efficiently and maintain reliable performance.

(c) Why is the series arrangement not used for domestic circuits?

Series arrangements are not used in domestic electric circuits because in a series circuit, the same current flows through all appliances, meaning if one appliance fails or is switched off, the entire circuit stops working. Also, electrical devices connected in series may receive different voltages, leading to improper functioning or damage. Parallel connections ensure that each device operates independently at the full supply voltage, providing safer and more reliable operation.

(d) How does the resistance of a wire vary with its area of cross-section?

The resistance of a wire is inversely proportional to its cross-sectional area. This means that if the area of cross-section increases, the resistance decreases. Mathematically, resistance \(R\) is given by $$\boxed{R = \rho \frac{l}{A}} $$ where,
\(\rho\) : is the resistivity,
\(l\) : is the length, and
\(A\) : is the area of cross-section.
Thus, doubling the area halves the resistance, making it easier for current to flow.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Copper and aluminium wires are commonly used for electrical transmission because both metals have high electrical conductivity, allowing efficient flow of electric current with minimal energy loss. Copper has excellent conductivity and mechanical strength, while aluminium, though less conductive than copper, is lighter and less expensive. Aluminium’s low density makes it easy to handle and reduces structural support needs, making it ideal for overhead transmission lines. Both metals also resist corrosion, contributing to the durability of electrical networks.