Question

An ammeter and voltmeter are connected in series to a 11 V battery. When a resistance is added in parallel with voltmeter, the voltmeter reading reduces to one-fourth and ammeter reading becomes three times as its initial value. Then

• A. The voltmeter reading after the connection of the resistance is 2 V
• B. If the resistance of the ammeter is 3 $\Omega$, then resistance of the voltmeter is 8 $\Omega$
• C. If resistance of the ammeter is 3 $\Omega$, then resistance of the resistor which is added in parallel to the voltmeter is $\frac{8}{11}$ $\Omega$
• D. Voltage drop across ammeter after the connection of resistance is 9 V

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

Let $E$ be the EMF of the battery, $R_A$ be the resistance of the ammeter, and $R_V$ be the resistance of the voltmeter. Let $R$ be the resistance added in parallel with the voltmeter.

Scenario 1: Initial setup

The ammeter and voltmeter are connected in series. The total resistance in the circuit is $R_{total1} = R_A + R_V$. The initial current (ammeter reading) is $I_1 = \frac{E}{R_A + R_V}$. The initial voltmeter reading is $V_1 = I_1 R_V = \frac{E R_V}{R_A + R_V}$.

Given $E = 11$ V. So, $I_1 = \frac{11}{R_A + R_V}$ and $V_1 = \frac{11 R_V}{R_A + R_V}$.

Scenario 2: After adding a resistance

A resistance $R$ is added in parallel with the voltmeter. The equivalent resistance of the parallel combination of $R_V$ and $R$ is $R_P = \frac{R_V R}{R_V + R}$. The new total resistance in the circuit is $R_{total2} = R_A + R_P = R_A + \frac{R_V R}{R_V + R}$. The new current (ammeter reading) is $I_2 = \frac{E}{R_A + R_P} = \frac{11}{R_A + \frac{R_V R}{R_V + R}}$. The new voltmeter reading is $V_2 = I_2 R_P = I_2 \frac{R_V R}{R_V + R}$.

From the problem statement:

1. The voltmeter reading reduces to one-fourth: $V_2 = \frac{V_1}{4}$.
2. The ammeter reading becomes three times its initial value: $I_2 = 3 I_1$.

Let's use these two conditions to find relationships between $R_A, R_V, R$.

From $V_2 = \frac{V_1}{4}$:

$I_2 R_P = \frac{1}{4} (I_1 R_V)$. Substitute $I_2 = 3 I_1$: $3 I_1 R_P = \frac{1}{4} I_1 R_V$. Since $I_1 \neq 0$, we can cancel $I_1$: $3 R_P = \frac{1}{4} R_V \implies R_P = \frac{R_V}{12}$. Now substitute $R_P = \frac{R_V R}{R_V + R}$: $\frac{R_V R}{R_V + R} = \frac{R_V}{12}$. Since $R_V \neq 0$, we can cancel $R_V$: $\frac{R}{R_V + R} = \frac{1}{12}$. $12 R = R_V + R \implies 11 R = R_V$. So, $R_V = 11 R$. This is our first key relationship.

From $I_2 = 3 I_1$:

$\frac{E}{R_A + R_P} = 3 \frac{E}{R_A + R_V}$. Cancel $E$: $\frac{1}{R_A + R_P} = \frac{3}{R_A + R_V}$. $R_A + R_V = 3 (R_A + R_P)$. Substitute $R_P = \frac{R_V}{12}$: $R_A + R_V = 3 \left(R_A + \frac{R_V}{12}\right)$. $R_A + R_V = 3 R_A + \frac{3 R_V}{12}$. $R_A + R_V = 3 R_A + \frac{R_V}{4}$. $R_V - \frac{R_V}{4} = 3 R_A - R_A$. $\frac{3 R_V}{4} = 2 R_A \implies R_V = \frac{8}{3} R_A$. This is our second key relationship.

Now we have two relationships:

1. $R_V = 11 R$
2. $R_V = \frac{8}{3} R_A$

From these, we can also express $R_A$ in terms of $R$:

$11 R = \frac{8}{3} R_A \implies R_A = \frac{33}{8} R$.

Now let's evaluate each option:

A. The voltmeter reading after the connection of the resistance is 2 V

First, calculate $V_1$: $V_1 = \frac{E R_V}{R_A + R_V}$. Substitute $R_V = \frac{8}{3} R_A$: $V_1 = \frac{11 \times (\frac{8}{3} R_A)}{R_A + \frac{8}{3} R_A} = \frac{11 \times \frac{8}{3} R_A}{\frac{11}{3} R_A} = \frac{11 \times 8}{11} = 8$ V. Since $V_2 = \frac{V_1}{4}$: $V_2 = \frac{8 \text{ V}}{4} = 2$ V. Option A is correct.

B. If the resistance of the ammeter is 3 $\Omega$, then resistance of the voltmeter is 8 $\Omega$

Given $R_A = 3 \, \Omega$. Using the relation $R_V = \frac{8}{3} R_A$: $R_V = \frac{8}{3} \times 3 \, \Omega = 8 \, \Omega$. Option B is correct.

C. If resistance of the ammeter is 3 $\Omega$, then resistance of the resistor which is added in parallel to the voltmeter is $\frac{8}{11}$ $\Omega$

Given $R_A = 3 \, \Omega$. From Option B, we found $R_V = 8 \, \Omega$. Using the relation $R_V = 11 R$: $8 \, \Omega = 11 R \implies R = \frac{8}{11} \, \Omega$. Option C is correct.

D. Voltage drop across ammeter after the connection of resistance is 9 V

The voltage drop across the ammeter after the connection of resistance is $V_{A2} = I_2 R_A$. We know $I_2 = 3 I_1$. $I_1 = \frac{E}{R_A + R_V}$. Using $R_A = 3 \, \Omega$ and $R_V = 8 \, \Omega$ (from options B and C): $I_1 = \frac{11 \text{ V}}{3 \, \Omega + 8 \, \Omega} = \frac{11 \text{ V}}{11 \, \Omega} = 1$ A. So, $I_2 = 3 I_1 = 3 \times 1 \text{ A} = 3$ A. The voltage drop across the ammeter is $V_{A2} = I_2 R_A = 3 \text{ A} \times 3 \, \Omega = 9$ V. Option D is correct.

All options (A, B, C, D) are correct.