Question

3. A battery of 10 volt carries 20,000 C of charge through a resistance of 20 Ω. The work done in 10 seconds is
(a) 2 × 103 joule
(b) 2 × 105joule
(c) 2 × 104 joule
(d) 2 × 102 joule

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Work\:Done=2\times 10^{5}\:J}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Potential \: difference(v) = 10 \: volt \\ \\ \tt: \implies Charge(Q) = 20000 \: c \\ \\ \tt: \implies Resistance( \Omega) = 20 \: \Omega \\ \\ \tt :\implies Time(t) = 10 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Work \: Done =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies I= \frac{Q}{t}\\ \\ \tt: \implies I = \frac{20000}{10} \\ \\ \green{\tt: \implies I= 2000 \: A} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Work \: done = Volt \times Current\times Time\\ \\ \tt: \implies Work \: Done = 10\times 2000 \times 10 \\ \\ \tt: \implies Work \: Done = 200000 \\ \\ \green{\tt: \implies Work \: Done = 2 \times {10}^{5}\:J }$