Two resistors each of resistance 2 Ohm are first connected in series and then in parallel across a potential difference of 12V. Find the ratio of the power consumed in series to that of power in parallel combination.
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potential difference=12V
for resistance in series,
Req= r1+r2+r3+.......+rn
Req= 2+2= 4 ohm
V=IR
12=I*4
I=12/4=3A
P=VI
power(P1)= 12*3=36
for parallel,
1/Req=1/r1 + 1/r2 + 1/r3 + ..... + 1/rn
1/Req = 1/2+1/2
=1 ohm
V=IR
12=I*1
I= 12A
P=VI
P2= 12*12 = 144
therefore ratio is P1/P2
= 36/144= 1/4
for resistance in series,
Req= r1+r2+r3+.......+rn
Req= 2+2= 4 ohm
V=IR
12=I*4
I=12/4=3A
P=VI
power(P1)= 12*3=36
for parallel,
1/Req=1/r1 + 1/r2 + 1/r3 + ..... + 1/rn
1/Req = 1/2+1/2
=1 ohm
V=IR
12=I*1
I= 12A
P=VI
P2= 12*12 = 144
therefore ratio is P1/P2
= 36/144= 1/4
AasthaSingh1:
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