The voltage V (in volts) across a circuit is given by Ohm\'s law: V=I*R, where I
ID: 3212933 • Letter: T
Question
The voltage V (in volts) across a circuit is given by Ohm's law: V=I*R, where I is the current (in amps) flowing through the circuit and R is the resistance (in ohms). If we place two circuits, with resistance R1 and R2, in parallel, then their combined resistance R is given by R= (R1*R2) / (R1+R2). Suppose the current is 2 amps and increasing at 10^-2 amp/sec and R1 is 3 ohms and increasing at 0.5 ohm/sec while R2 is 5 ohms and decreasing at 0.1 ohm/sec. Find the rate at which the voltage is changing (in volt/sec).Explanation / Answer
Rewrite 1/R = 1/R1 + 1/R2 in terms of R: R = [1/R1 + 1/R2]?¹. When R1 = 6 and R2 = 13, R = 78/19 ohm. Differentiate both sides with respect to time t: dR/dt = -[1/R1 + 1/R2]?² * [(-1/R1²) dR1/dt + (-1/R2²) dR2/dt] .........= [1/R1 + 1/R2]?² [(1/R1²) dR1/dt + (1/R2²) dR2/dt] Using the given information, we can solve for dR/dt: dR/dt = [1/4 + 1/6]?² [(1/6²) (0.5) + (1/13²) (-0.1)] = 1618/21125. Now, differentiate both sides of V = IR with respect to time: dV/dt = I dR/dt + R dI/dt
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