Following figures show different combinations of identical bulb(s) connected to identical battery(ies). Which option is correct regarding the total power dissipated in the circuit?

Question

(A) P< Q< R< S (B) P< Q< R=S (C) R < Q < P < S (D) P > R > Q >S

Tardigrade Admin · Accepted Answer

Power consumed in each case is (P) P=(V2/3 R)

(Q)P=(V2/R / 3)=(3 V2/R)

(R)P=(V2/R)

(S)P=((2 V)2/R)=(4 V2/R)

So, order of increasing power consumption is P >R >Q >S.

Q. Following figures show different combinations of identical bulb(s) connected to identical battery(ies). Which option is correct regarding the total power dissipated in the circuit?

$P < Q < R < S$

$P < Q < R = S$

$R < Q < P < S$

$P > R > Q > S$

Power consumed in each case is
(P) $P = \frac{V ^{2}}{3 R}$

(Q) $P = \frac{V ^{2}}{R /3} = \frac{3 V ^{2}}{R}$

(R) $P = \frac{V ^{2}}{R}$

(S) $P = \frac{( 2 V ) ^{2}}{R} = \frac{4 V ^{2}}{R}$

So, order of increasing power consumption is
$P > R > Q > S$ .

Q. Following figures show different combinations of identical bulb(s) connected to identical battery(ies). Which option is correct regarding the total power dissipated in the circuit?

P<Q<R<S

P<Q<R=S

R<Q<P<S

P>R>Q>S

Power consumed in each case is (P) P=3RV2​ (Q)P=R/3V2​=R3V2​ (R)P=RV2​ (S)P=R(2V)2​=R4V2​ So, order of increasing power consumption is P>R>Q>S.

$P < Q < R < S$

$P < Q < R = S$

$R < Q < P < S$

$P > R > Q > S$

Power consumed in each case is
(P) $P = \frac{V ^{2}}{3 R}$

(Q) $P = \frac{V ^{2}}{R /3} = \frac{3 V ^{2}}{R}$

(R) $P = \frac{V ^{2}}{R}$

(S) $P = \frac{( 2 V ) ^{2}}{R} = \frac{4 V ^{2}}{R}$

So, order of increasing power consumption is
$P > R > Q > S$ .