Problem 20 - A Complete Resource Book in Physics for JEE Main

Question

The resistances \(500 \Omega\) and \(1000 \Omega\) are connected in series with a battery of \(1.5 \mathrm{~V}\). The voltage across the \(1000 \Omega\) resistance is measured by a voltmeter having a resistance of \(1000 \Omega\). The reading in the voltmeter would be (A) \(1.5 \mathrm{~V}\) (B) \(1.0 \mathrm{~V}\) (C) \(0.75 \mathrm{~V}\) (D) \(0.5 \mathrm{~V}\)

Step-by-Step Solution

Answer

The reading on the voltmeter would be (C) $0.75 \mathrm{~V}$. This is determined by first calculating the equivalent resistance of the series circuit, then using Ohm's law to find the current. We then calculate the voltage across the 1000 Ω resistor. When the 1000 Ω voltmeter is added, a new equivalent resistance is found and a new total current value is calculated. Using the calculated current, the voltage across the 1000 Ω resistor with the voltmeter connected is 0.75 V.

Step 1: Calculate the equivalent resistance of the circuit

Firstly, we will find the equivalent resistance of the circuit without the voltmeter connected. The resistances are connected in series, so their resistances are added. \(R_{eq} = R_1 + R_2 = 500 \Omega + 1000 \Omega = 1500 \Omega\)

Step 2: Use Ohm's law to find the total current passing through the circuit

We are given the battery voltage V = 1.5 V. Using Ohm's law, we will calculate the total circuit current: \(I = \frac{V}{R_{eq}} = \frac{1.5 \mathrm{~V}}{1500 \Omega} = 0.001 \mathrm{~A}\)

Step 3: Determine the voltage across the 1000 Ω resistor (without voltmeter)

The voltage across the 1000 Ω resistor without the voltmeter can be found using the formula: \(V_{R_2} = I \times R_2 = 0.001 \mathrm{~A} \times 1000 \Omega = 1 \mathrm{~V}\)

Step 4: Calculate the new equivalent resistance with the voltmeter connected

Now, let's find the equivalent resistance when the voltmeter with resistance 1000 Ω is connected across the 1000 Ω resistor. The voltmeter is connected in parallel to the 1000 Ω resistor. To calculate the equivalent resistance we add 500 Ω (unchanged resistor) and the parallel combination of the voltmeter and the 1000 Ω resistor. \[ \frac{1}{R_{eq}} = \frac{1}{R_2} + \frac{1}{R_{V}}\] \[R_{eq} = R_1+\frac{R_2R_V}{R_2+R_V} = 500 \Omega + \frac{1000\Omega \times 1000 \Omega}{1000 \Omega+1000 \Omega} = 500\Omega + 500\Omega = 1000\Omega\]

Step 5: Calculate the total current passing through the circuit with voltmeter connected

Using Ohm's law for the modified circuit, the total circuit current can be calculated as: \(I' = \frac{V}{R_{eq}} = \frac{1.5 \mathrm{~V}}{1000 \Omega} = 0.0015 \mathrm{~A}\)

Step 6: Use the current divider rule to find the voltage across the 1000 Ω resistor with the voltmeter connected

The current through the 1000 Ω resistor and the voltmeter can be found using the current divider rule: \[I_{R_2} = I' \times \frac{R_{V}}{R_2 + R_{V}} = 0.0015 \mathrm{~A} \times \frac{1000\Omega}{(1000\Omega+1000\Omega)} = 0.00075 \mathrm{~A}\] Finally, we can find the voltage across the 1000 Ω resistor with the voltmeter connected: \(V'_{R_2} = I_{R_2} \times R_2 = 0.00075 \mathrm{~A} \times 1000 \Omega = 0.75 \mathrm{~V}\)

Step 7: Compare the calculated value with the given options

We have found the voltmeter reading to be 0.75 V. Comparing the calculated value with the given options, we find that option (C) 0.75 V is the correct answer.