A cell of e.m.f 2 V and internal resistance 0.1 Ohm is connected to a 3.9 ohm external resistance.What will be the p.d across the terminals of the cell?

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Solution

R= 3.9 ohm
r= 0.1 ohm
E= 2V
E= I(R+r)
2=I(3.9+0.1)
I=1/2= 0.5A
Potential difference between the terminals of a battery = Terminal voltage
Since E= V +v
Where V= terminal voltage
v= voltage drop
and v= Ir =0.5 x 0.1 =0.05 ohm
V= E-v = 2 – 0.05 = 1.95 V

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A cell of e.m.f 2 V and internal resistance 0.1 Ohm is connected to a 3.9 ohm external resistance.What will be the p.d across the terminals of the cell?

R= 3.9 ohmr= 0.1 ohmE= 2VE= I(R+r)2=I(3.9+0.1)I=1/2= 0.5APotential difference between the terminals of a battery = Terminal voltageSince E= V +vWhere V= terminal voltagev= voltage dropand v= Ir =0.5 x 0.1 =0.05 ohmV= E-v = 2 – 0.05 = 1.95 V

R= 3.9 ohm
r= 0.1 ohm
E= 2V
E= I(R+r)
2=I(3.9+0.1)
I=1/2= 0.5A
Potential difference between the terminals of a battery = Terminal voltage
Since E= V +v
Where V= terminal voltage
v= voltage drop
and v= Ir =0.5 x 0.1 =0.05 ohm
V= E-v = 2 – 0.05 = 1.95 V