The voltage V (in volts) across a circuit is given by Ohm\'s Law V=IR Where I is

Question

The voltage V (in volts) across a circuit is given by Ohm's Law V=IR Where I is the current (in amps flowing through the circuit, and R is the resistance (in Ohm's). If we place two circuits with resistance R1 and R2 in parallel, then their combinded resistance is given by 1/R=1/R1 + 1/R2

Supose

(a) the current is 2 amps and is increasing at 10^(-2) amps/sec

(b) R1 is 3 ohms and is increasing at 0.5 ohms/sec

(c) R2 is 5 ohms and is decreasing at 0.1 ohms/sec

Calculate the rate at which the voltage is changing

Explanation / Answer

V = IR =>dV/dt = (R*(dI/dt)) + (I*(dR/dt))............(1) We have to find dV/dt given dI/dt = 10-2amps/sec. (1/R) = (1/R1) + (1/R2) =>(-1/R2)*dR/dt = ((-1/R12)*dR1/dt) + ((-1/R22)*dR2/dt).................(2) R1 = 3ohms, R2 = 5ohms, R = (R1*R2)/(R1+R2) = 15/8 ohms dR1/dt = 0.5ohms/sec, dR2/dt=-0.1ohms/sec Substituting above values in (2) we get, dR/dt = 0.18125ohms/sec Substituting above values in (1) we get, dV/dt = 0.38125Volts/sec

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