Energy Stored in a Magnetic Field — Question 4

Question 4

Series discharge resistors divide the initial magnetic energy in proportion to resistance

An ideal .400-H inductor initially carries 2.00 A and discharges through series resistors R1=3.00 ohms and R2=5.00 ohms. There is no source and no other loss.

1. Find the total heat and its share in each resistor.

2. Find the fraction of the initial energy remaining after time L/(R1+R2).

Question 4 - Solution

Series discharge resistors divide the initial magnetic energy in proportion to resistance

(a) Method and calculation

$$U_0=\tfrac12LI_0^2=\tfrac12(.400)(2.00)^2=.800\,J.$$$$H_j=\int i^2R_jdt=\frac{R_j}{R_1+R_2}U_0\Rightarrow\boxed{H_1=.300\,J,\ H_2=.500\,J}.$$

(b) Interpretation and further calculation

$$\tau=L/(R_1+R_2)=.0500\,s.$$$$\boxed{U(\tau)/U_0=e^{-2}=.135335}.$$

Physical check

The energy shares depend on resistance ratios because the same time-dependent current flows through both resistors.