CBSE Class 12 · 2015 · Delhi · Current ElectricityShort Answer

A cell of EMF

E

and internal resistance

r

is connected across a variable load

R

. Draw the plots of
(i) the terminal voltage

V

and
(ii) the current

I

versus

R

. It is found that when

R = 4 Ω

the current is

1 A

, and when

R = 9 Ω

the current is

0.5 A

. Find the EMF and the internal resistance.

Answer: $E = 5 V$ , $r = 1 Ω$ .

Solution

Using

E = I (R + r)

: for

R = 4 Ω

E = 1 (4 + r)

; for

R = 9 Ω

E = 0.5 (9 + r)

. Equating,

4 + r = 0.5 (9 + r) \Rightarrow 4 + r = 4.5 + 0.5 r \Rightarrow 0.5 r = 0.5 \Rightarrow r = 1 Ω

, and then

E = 4 + 1 = 5 V .

CBSE Class 12 Board · 2015 · Delhi · Subjective