Figure shows an automobile circuit. How much power (in watt) is dissipated by the automobile circuit when switches A, B, C and D are all closed.

Question

Figure shows an automobile circuit. How much power (in watt) is dissipated by the automobile circuit when switches A, B, C and D are all closed. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

All the elements of circuit are in parallel arrangement (1/ R eq )=(1/40) × (1/40) × (1/40) × (1/40) × (1/20) × (1/20) =(4/40) × (2/40) R eq =5 Ω Power = V 2 / R =20 W

Q. Figure shows an automobile circuit. How much power (in watt) is dissipated by the automobile circuit when switches $A, B, C$ and $D$ are all closed.

All the elements of circuit are in parallel arrangement
$\frac{1}{R _{e q}} = \frac{1}{40} \times \frac{1}{40} \times \frac{1}{40} \times \frac{1}{40} \times \frac{1}{20} \times \frac{1}{20}$
$= \frac{4}{40} \times \frac{2}{40}$
$R_{e q} = 5 Ω$
Power $= V^{2} / R = 20 W$

Q. Figure shows an automobile circuit. How much power (in watt) is dissipated by the automobile circuit when switches A,B,C and D are all closed.

All the elements of circuit are in parallel arrangement Req​1​=401​×401​×401​×401​×201​×201​ =404​×402​ Req​=5Ω Power =V2/R=20W

Q. Figure shows an automobile circuit. How much power (in watt) is dissipated by the automobile circuit when switches $A, B, C$ and $D$ are all closed.

All the elements of circuit are in parallel arrangement
$\frac{1}{R _{e q}} = \frac{1}{40} \times \frac{1}{40} \times \frac{1}{40} \times \frac{1}{40} \times \frac{1}{20} \times \frac{1}{20}$
$= \frac{4}{40} \times \frac{2}{40}$
$R_{e q} = 5 Ω$
Power $= V^{2} / R = 20 W$