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Question

In an electrical circuit two resistors of 2 $$\Omega$$ and 4 $$\Omega$$ respectively are connected in series to a 6 V battery. The heat dissipated by the 4 $$\Omega$$ resistor in 5 s will be


A
5 J
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B
10 J
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C
20 J
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D
30 J
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Solution

The correct option is C 20 J
Hint : Use the mathematical expression of Joule's Law to evaluate heat dissipated.

Step 1: Find the equivalent resistance of the Circuit.

                    Given: Two resistors $$R_{1}$$ &  $$R_{2}$$ are connected in series.
                     So, equivalent resistance can be given as:
                    $$R_{net}$$ = $$R_{1}\ + \ $$$$R_{2}$$ 
                    $$R_{net}$$ = $$2\ + \ $$$$4 =$$ $$6 \Omega$$

Step 2 : Find the current through $$4 \Omega$$ resistor

                     Since both resistors are connected in series so current through 
                     each resistor is same as the current through the circuit.
                     $$Current = \dfrac{P.D.}{Net\ resistance}$$
                     $$ I =\dfrac{6}{6}=1A$$

 Step 3: Heat dissipated through $$4 \Omega$$ resistor can be given as :

                       $$H= I^2Rt$$
                       where, $$H =$$ Heat dissipated, $$R$$= Resistance and $$t =$$ time in seconds 
                       So, $$H = 1^2×4×5=20J$$

Final Step : Heat dissipated through $$4 \Omega$$ resistor in $$5\ sec$$ is  $$20\ J$$. Option C is correct. 


Physics

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