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

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5 J

10 J

20 J

30 J

answer is C.

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Detailed Solution

If in an electrical circuit, two resistors of

2 Ω

and

4 Ω

respectively are connected in series to a

6 V

battery, then the heat dissipated by the

4 Ω

resistor in

5 s

20 J

.
Two resistances of

2 Ω

and

4 Ω

are given. Also, these resistances are connected in series.
It is known that “when resistors are connected in series, the equivalent resistance of the combination is given by the sum of the individual resistances.” Mathematically,

R = R_{1} + R_{2} + \dots + R_{n}

Where

R

is equal to the equivalent resistance and

R_{1}, R_{2}, \dots, R_{n}

are the individual resistances.
On substituting the given values, we get

R_{s} = 2 + 4

\Rightarrow R_{s} = 6 Ω

Using Ohm’s law, the current can be calculated as,

I = \frac{V}{R_{s}}

\Rightarrow I = \frac{6}{6}

\Rightarrow I = 1 A

Therefore, heat dissipation in the

4 Ω

resistor is equal to

H = I^{2} Rt

\Rightarrow H = 1^{2} \times 4 \times 5

\Rightarrow H = 20 J

Hence, the heat dissipation is equal to

20

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