Question

PHYSICS
A cell supplies 0.6A to a 2ohm coil
And 0.3A to another coil of 8ohm. (Both coils simultaneously)
Find the EMF(p.d.) and internal resistance of the entire circuit.

Answer 1

Here, I1 = 0.6 A, R₁ = 2 Ω, I₂ = 0.3 A, R₂ = 8 Ω, r = ?

Using I = E / (R + r)

0.6 = E / (2 + r) -------------- (i)

and 0.3 = E / (8 + r) --------------- (ii)

From (i) and (ii) we get E = 0.6 (2 + r) = 0.3 (8 + r)

Or, 0.3 r = 1.2 => r = 4 ohm.

Substituting the value of r in (i) we get

0.6 = E / (2 + 4) => E = 0.6 (2 + 4) = 3.6 V.
OR

EMF is denoted with "E"

and internal resisitance with "r"

we know,

V=IR

for current 0.6 A and resistance 2ohm

V=0.6×2=1.2volt

now,

for current 0.3 A and resistance 8ohm

V=0.3×8= 2.4 volt

as we know,

V=E-Ir

so,

for V=1.2 volt

1.2=E-0.6×r......( 1 )

and

for V=2.4 volt

2.4=E-0.3 ×r........( 2 )

solving equation 1 & 2

we get

internal resistance( r ) = 4 ohm

and

EMF of cell ( E ) = 3.6 volt