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
B
10 J
C
20 J
D
30 J

Solution

## The correct option is C 20 JHint : 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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