Current Electricity

Practice Questions:

Question 51: difficult

Three 10Ω, 2 W resistors are connected as in Fig. The maximum possible voltage between points A and B without exceeding the power dissipation limits of any of the resistors is:

1. 5√3 V

2. 3√5 V

3. 15 V

4. 5/3 V

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Question 52: moderate

Two light bulbs shown in the circuit have ratings A(24 V, 24 W) and B (24 V and 36 W) as shown. When the switch is closed.

1. The intensity of light bulb A increases

2. The intensity of light bulbs both A and B remains same

3. The intensity of light bulb B increases

4. The intensity of light bulb B decreases

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Question 53: moderate

The three resistances A, B and C have values 3R, 6R and R respectively. When some potential difference is applied across the network, the thermal powers dissipated by A, B and C are in the ratio :

1. 2 : 3 : 4

2. 2 : 4 : 3

3. 4 : 2 : 3

4. 3 : 2 : 4

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Question 54: moderate

An electric bulb rated for 500 watts at 100 volts is used in a circuit having a 200-volt supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 watts is :

1. 10 Ω

2. 20 Ω

3. 50 Ω

4. 100 Ω

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To find the resistance

$R$

that must be placed in series with the bulb, let's analyze the problem step by step.

Given:

Power of the bulb ( $P$

) = 500 W,
Voltage rating of the bulb ( $V_b$

) = 100 V,
Supply voltage ( $V_s$

) = 200 V.

Step 1: Resistance of the bulb

The resistance of the bulb (

$R_b$

) can be calculated using the formula:

$R_b = \frac{V_b^2}{P}.$

Substitute the values:

$R_b = \frac{100^2}{500} = \frac{10000}{500} = 20 \, \Omega.$

Step 2: Total current through the circuit

The bulb is rated to draw 500 W at 100 V. Thus, the current through the bulb is:

$I = \frac{P}{V_b}.$

Substitute the values:

$I = \frac{500}{100} = 5 \, \text{A}.$

Step 3: Voltage drop across the series resistor

The total supply voltage is 200 V, and the bulb operates at 100 V. Therefore, the voltage drop across the series resistor

$R$

is:

$V_R = V_s - V_b.$

Substitute the values:

$V_R = 200 - 100 = 100 \, \text{V}.$

Step 4: Resistance of the series resistor

Using Ohm's law, the resistance of the series resistor is:

$R = \frac{V_R}{I}.$

Substitute the values:

$R = \frac{100}{5} = 20 \, \Omega.$

Final Answer:

The resistance that must be placed in series with the bulb is:

$\boxed{20 \, \Omega}.$

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Question 55: moderate

An ammeter is to be constructed which can read currents upto 2.0 A. If the coil has a resistance of 25 Ω and takes 1 mA for full-scale deflection, what should be the resistance of the shunt used?

1. \[1.25\times 10^{-2}\Omega\]

2. \[2.5\times 10^{-2}\Omega\]

3. \[0.5\times 10^{-2}\Omega\]

4. \[10^{-2}\Omega\]

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To construct an ammeter that can read currents up to

$2.0 \, \text{A}$

, we need to calculate the resistance of the shunt. Here's the step-by-step calculation:

Full-scale current through the coil:
The coil takes $1 \, \text{mA} = 10^{-3} \, \text{A}$

for full-scale deflection.
Remaining current through the shunt:
When the total current is $2.0 \, \text{A}$

, the current through the shunt is:

$I_s = I_{\text{total}} - I_c = 2.0 - 10^{-3} = 1.999 \, \text{A}.$
Voltage across the coil:
The resistance of the coil is $25 \, \Omega$

. Using Ohm's law, the voltage across the coil is:

$V_c = I_c R_c = (10^{-3})(25) = 0.025 \, \text{V}.$
Resistance of the shunt:
The voltage across the shunt must equal the voltage across the coil: $V_s = V_c = 0.025 \, \text{V}.$ Using Ohm's law for the shunt:

$R_s = \frac{V_s}{I_s} = \frac{0.025}{1.999} \approx 1.25 \times 10^{-2} \, \Omega.$

Thus, the resistance of the shunt is:

$\boxed{1.25 \times 10^{-2} \, \Omega.}$

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