Question

) The internal resistance of a cell of emf 3V is 0.2. It is connected to a reistance of 0.8omega . The
voltage across the cell will be​

Answer 1

Answer:

Given: E=2V, r=1.2Ω, R= (external resistance)

Let 4.5Ω and 9Ω connected in parallel, then equivalent resistance

R

p

1

=

4.5

1

+

9

1

R

p

1

=

9

2+1

=

9

3

=

3

1

R

p

=3Ω

Now, 0.8Ω and R

p

resistances are in series. Then total resistance

R=3+0.8=3.8Ω

(i) Current in 0.8Ω resistance I=

R+r

E

=

3.8+1.2

2

=

5

2

=0.4A

(ii) Terminal Voltage (V)=E−Ir=2−(0.4×1.2)

=2−0.48

=1.52 V