Question

The value of resistance $R$ is being measured with the potentiometer set up in two steps shown below. Step-I : $E_s = 10$ V, $r = 1 \Omega$ and zero deflection is observed at length $PJ = 80$ cm. Step-II : $E_s = 5$ V, $r = 2 \Omega$ and zero deflection is observed at length $PJ = 30$ cm. Then

• A. The value of $R$ is $1 \Omega$
• B. The value of $R$ is $2 \Omega$
• C. Current through $R$ in Step-I is $\frac{10}{3} A$
• D. Current through $R$ in Step-II is $\frac{5}{4} A$

Answer 1

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

The value of $R$ is $2 \Omega$, Current through $R$ in Step-I is $\frac{10}{3} A$, Current through $R$ in Step-II is $\frac{5}{4} A$

Answer 2

The problem involves a potentiometer circuit used to determine an unknown resistance $R$. The principle of a potentiometer states that at null deflection, the potential difference across the balancing length of the potentiometer wire is equal to the potential difference across the component in the secondary circuit.

The secondary circuit consists of a cell $E_s$, an internal resistance $r$, and the unknown resistance $R$. The current flowing through this secondary circuit is given by Ohm's law: $I_s = \frac{E_s}{R+r}$

The potential difference across the unknown resistance $R$ is: $V_R = I_s \times R = \frac{E_s R}{R+r}$

At null deflection, this potential difference $V_R$ is balanced by the potential drop across the balancing length $L$ of the potentiometer wire. If $k$ is the potential gradient of the potentiometer wire (potential drop per unit length), then: $k \times L = V_R$ So, $k \times L = \frac{E_s R}{R+r}$

We are given two steps:

Step-I: $E_s = 10$ V $r = 1 \Omega$ $L = 80$ cm Substituting these values into the equation: $k \times 80 = \frac{10 \times R}{R+1}$ (Equation 1)

Step-II: $E_s = 5$ V $r = 2 \Omega$ $L = 30$ cm Substituting these values into the equation: $k \times 30 = \frac{5 \times R}{R+2}$ (Equation 2)

To find $R$, we can divide Equation 1 by Equation 2: $\frac{k \times 80}{k \times 30} = \frac{\frac{10R}{R+1}}{\frac{5R}{R+2}}$ $\frac{8}{3} = \frac{10R}{R+1} \times \frac{R+2}{5R}$ The $R$ in the numerator and denominator on the right side cancels out, and $10/5 = 2$: $\frac{8}{3} = \frac{2(R+2)}{R+1}$ Now, cross-multiply: $8(R+1) = 3 \times 2(R+2)$ $8R + 8 = 6R + 12$ $8R - 6R = 12 - 8$ $2R = 4$ $R = 2 \Omega$

So, option B is correct.

Now, let's calculate the currents through $R$ for both steps:

Current through $R$ in Step-I ($I_{s1}$): $I_{s1} = \frac{E_s}{R+r} = \frac{10 \text{ V}}{2 \Omega + 1 \Omega} = \frac{10}{3}$ A So, option C is correct.

Current through $R$ in Step-II ($I_{s2}$): $I_{s2} = \frac{E_s}{R+r} = \frac{5 \text{ V}}{2 \Omega + 2 \Omega} = \frac{5}{4}$ A So, option D is correct.

Option A states $R = 1 \Omega$, which is incorrect.

Thus, options B, C, and D are all correct.