Question

The resistances of the four arms $P,Q, R$ and $S$ in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be :

• A. 2.0 A
• B. 1.0 A
• C. 0.2 A
• D. 0.1 A

Answer 1

The Correct Option is C

Total resistance of Wheatstone bridge $ = \frac{(40)(120)}{40 + 120} = 30 \Omega $ Current through cell = $\frac{7 V}{( 5 +30) \Omega } = \frac{1}{5} A = 0.2 \, A $

Answer 2

Wheatstone bridge is used for the measurement of the unknown resistance connected in a circuit. It’s made of 4 resistors of which two resistors are known resistors, one resistor which can be varied, and the unknown resistor. It is also made of a galvanometer. Refer to the connections from the figure.

 

It functions based on the principle that when the value of resistors follows the relation

\(\frac{P}{Q}\)=\(\frac{R}{S}\)

the voltages at points b and d become equal and the current in the galvanometer G becomes zero.

In this question we have

P=10Ω

Q=30Ω

R=30Ω

S=90Ω

Therefore,

\(\frac{P}{Q}\) = \(\frac{10}{30}\) = \(\frac{R}{S}\) = \(\frac{30}{90}\) =\(\frac{1}{3}\)

Thus, no current flows in the galvanometer. And the final circuit looks like this.

We know,

For series ⇒ Req = R₁+R₂

For parallel ⇒ Req = \(\frac{R_1 R_2}{R_1+R_2}\)

The equivalent resistance of the circuit is calculated as follows:

 

R_{eq_up} = P+Q=40Ω

R_{eq_down} = R+S=120Ω

Req = \(\frac{R_ equp R_ eqdown}{R_equp + R_eqdown}\) = \(\frac{40\times120}{40+160}\) = \(\frac{4800}{160}\) = 30Ω

From the relation between voltage, current, and resistance we can say,

V=IR

The current is measured as,

I=\(\frac{V}{(R)battery+Req}\)=\(\frac{7}{30+5}\)=\(\frac{7}{35}\)=0.2A

Therefore, the answer to this question is option C. 

0.2A