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Question:

When two identical batteries of internal resistance 1Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J2, then the value of R in Ω is:

Solution:

Current I1 = (2E)/(2r + R)
Heat in situation 1 ⇒ J1 = I1²Rt ⇒ J1 = ((2E)/(2r + R))²Rt
Similarly I2 = (E/((r/2) + R))
Heat in situation 2 ⇒ J2 = I2²Rt ⇒ J2 = (E/((r/2) + R))²Rt
⇒ J1 = 2.25J2 (Given)
((2E)/(2r + R))² = 9/4 × (E/((r/2) + R))²
4(r + 2R)² = 9(2r + R)²
2(r + 2R) = 3(2r + R)
2r + 4R = 6r + 3R
R = 4r = 4Ω