Current Electricity

JEE Main level, single correct, medium

275 of 380

Question 275 of 380

A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances (R1=R2=R3=R4)\left(R_{1}=R_{2}=R_{3}=R_{4}\right). When R3R_{3} resistance is heated to some temperature, its resistance value has gone up by 10%10 \%. The potential difference (Va−Vb)\left(V_{a}-V_{b}\right) (after R3R_{3} is heated) is ____\_\_\_\_ V.

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Answer: (B) 0.95

Initially, the Wheatstone bridge is balanced with R1=R2=R3=R4=RR_1 = R_2 = R_3 = R_4 = R.
Since the bridge is balanced, the potential difference across the bridge is zero: Va−Vb=0V_a - V_b = 0 V.
When R3R_3 is heated and its resistance increases by 10%: R3′=1.1RR_3' = 1.1R
The voltage at node aa (between R1R_1 and R2R_2): Va=40×R2R1+R2=40×RR+R=40×12=20V_a = 40 \times \frac{R_2}{R_1 + R_2} = 40 \times \frac{R}{R + R} = 40 \times \frac{1}{2} = 20 V
The voltage at node bb (between R3′R_3' and R4R_4): Vb=40×R4R3′+R4=40×R1.1R+R=40×12.1≈19.05V_b = 40 \times \frac{R_4}{R_3' + R_4} = 40 \times \frac{R}{1.1R + R} = 40 \times \frac{1}{2.1} \approx 19.05 V
Therefore: Va−Vb=20−19.05=0.95V_a - V_b = 20 - 19.05 = 0.95 V

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