Question

A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor

Answer 1

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

$V(t) = \epsilon. e^{-t.\frac{L}{R} }$

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

$0.8*19 = 19. e^{-4.\frac{22}{R} }$

$0.8 = e^{-\frac{88}{R} }$

$ln(0.8) = ln(e^{-\frac{88}{R} })$

$ln(0.8) = -\frac{88}{R}$

$R = -\frac{88}{ln(0.8)}$

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.