In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

Question

In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be (A)  200 Ω  (B)  100 Ω  (C)  500 Ω  (D)  1000 Ω

Tardigrade Admin · Accepted Answer

The galvanometer shows zero deflection ie, current through XY is zero.
As a result potential drop across R is 2 V, circuit can be redrawn as

I = \frac{12}{500 + R}

Voltage across

R, V = I R

\Rightarrow

2 = \frac{12}{500 + R} \times R

\Rightarrow

1000 + 2 R = 12 R \Rightarrow R = 100 Ω

Q. In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

$200 Ω$

$100 Ω$

$500 Ω$

$1000 Ω$

The galvanometer shows zero deflection ie, current through XY is zero.
As a result potential drop across R is 2 V, circuit can be redrawn as
$I = \frac{12}{500 + R}$
Voltage across $R, V = I R$
$\Rightarrow$ $2 = \frac{12}{500 + R} \times R$
$\Rightarrow$ $1000 + 2 R = 12 R \Rightarrow R = 100 Ω$

Q. In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

200Ω

100Ω

500Ω

1000Ω

The galvanometer shows zero deflection ie, current through XY is zero. As a result potential drop across R is 2 V, circuit can be redrawn as I=500+R12​ Voltage across R,V=IR ⇒ 2=500+R12​×R ⇒ 1000+2R=12R⇒R=100Ω

$200 Ω$

$100 Ω$

$500 Ω$

$1000 Ω$

The galvanometer shows zero deflection ie, current through XY is zero.
As a result potential drop across R is 2 V, circuit can be redrawn as
$I = \frac{12}{500 + R}$
Voltage across $R, V = I R$
$\Rightarrow$ $2 = \frac{12}{500 + R} \times R$
$\Rightarrow$ $1000 + 2 R = 12 R \Rightarrow R = 100 Ω$