Question

Two resistance R1=10 ohm and R2 =20 ohm are connected in series with a battery of emf E=10 v
then potential access resistance R1 is
a) 3.333v
b)6.666v
c) 0.333v
d) 0.666v

Answer 1

Answer:

Mark it as BRAINLIEST

Explanation:

The circuit current

I = 12/(10+20)

I = 12/30

I =0.4 Amps.

Power dissiparion in 10- ohm resistor

P1= I^2R1

P1 = 0.4 x0. 4 x10

P1= 1.6 watts

Power dissiparion in 20- ohm resistor

P2= I^2R2

P2 = 0.4 x0. 4 x20

P2= 3.2 watts