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Question:
In the circuit shown in figure the heat produced in the $5\Omega$ resistor due to the current flowing through it is 10 cal/s.
The heat generated in the $4\Omega$ resistor is
In the circuit shown in figure the heat produced in the $5\Omega$ resistor due to the current flowing through it is 10 cal/s.
The heat generated in the $4\Omega$ resistor is
Updated On: Jun 23, 2023
- 1 cal/s
- 2 cal/s
- 3 cal/s
- 4 cal/s
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The Correct Option is B
Solution and Explanation
Since, resistance in upper branch of the circuit is twice the
resistance in lower branch. Hence, current there will be half
$P_4=(i/2)^2(4) \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (p=i^2R)$
$P_5 = (i)^2(5)$
or $\, \, \, \, \, \, \, \, \, \, \, \frac{P_4}{P_5} = \frac{1}{5}$
$P_4 = \frac{P_5}{5} = \frac{10}{5} = 2 \, cal/s$
$\therefore \, $Correct option is (b)
resistance in lower branch. Hence, current there will be half
$P_4=(i/2)^2(4) \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (p=i^2R)$
$P_5 = (i)^2(5)$
or $\, \, \, \, \, \, \, \, \, \, \, \frac{P_4}{P_5} = \frac{1}{5}$
$P_4 = \frac{P_5}{5} = \frac{10}{5} = 2 \, cal/s$
$\therefore \, $Correct option is (b)
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