In Fig. switch S is closed for a long time. At $t = 0$ , if it is opened, then

total heat produced in resistor

R

after opening the switch is

\frac{1}{2} \frac{L ε}{R^{2}}

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total heat produced in resistor

R_{1}

after opening the switch is

\frac{1}{2} \frac{L ε^{2}}{R^{2}} (\frac{R_{1}}{R_{1} + R_{2}})

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heat produced in resistor

R_{1}

after opening the switch is

\frac{1}{2} \frac{R_{2} L ε^{2}}{(R_{1} + R_{2}) R^{2}}

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no heat will be produced in

R_{1}

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Solution

The correct option is C heat produced in resistor $R_{1}$ after opening the switch is $\frac{1}{2} \frac{R_{2} L ε^{2}}{(R_{1} + R_{2}) R^{2}}$
Just before opening the switch, the current in the inductor is $\frac{E}{R}$ . Energy store in it = $\frac{1}{2} (\frac{L}{R})^{2}$
This energy will dissipate in resistors $R_{1}$ and $R_{2}$ in the ratio $\frac{1}{R_{1}}$ and $\frac{1}{R_{2}}$ .
heat produced in resistor $R_{1}$ after opening the switch is $\frac{1}{2} L (\frac{}{R})^{2} \frac{R_{2}}{R_{1} + R_{2}}$

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Similar questions

Q. In the circuit shown in figure,

E = 120 V, R_{1} = 3.0 Ω, R_{2} = 5.0 Ω

and

L = 0.200 H

. Switch

S

is closed at

t = 0

. The switch is closed for a long time and then opened. Just after the switch is opened. What is the potential difference

V_{a b}

across the resistor

R_{1}

Q. A simple circuit consisting of three resistors is shown above.

R_{1}

has twice the resistance of

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. If the switch remains open, determine the ratio of heat dissipated of

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