Question:
An electric circuit is given below. V1 = 1 V and
Resistance R = 1000 W.
A
B
V R 1
The current through the resistance R is very
close to 1 mA and the voltage across point A
and B, VAB = 1 V. Now the circuit is changed
to:
A
B
V R 1 V2
where value of V2 = 5V. The internal resistances
of both the batteries are 0.1 W. The current
through the resistance R is about:
An electric circuit is given below. V1 = 1 V and
Resistance R = 1000 W.
A
B
V R 1
The current through the resistance R is very
close to 1 mA and the voltage across point A
and B, VAB = 1 V. Now the circuit is changed
to:
A
B
V R 1 V2
where value of V2 = 5V. The internal resistances
of both the batteries are 0.1 W. The current
through the resistance R is about:
Options:
- 1.0 mA
- 1.2 mA
- 3.0 mA
- 5.0 mA
- 1.0 mA
- 1.2 mA
- 3.0 mA
- 5.0 mA
Correct Answer: C) 3.0 mA
Explanation:
Equivalent voltage = (e1*r1 + e2*r2)/(r1 + r2) = (1*0.1 + 5*0.1)/(0.1 + 0.1) = 0.6 / 0.2 = 3 V
From first circuit, we have i = e/(R+r). Value of r is very less and can be neglected as compared to R.
Now, i = e/R => 10^-3 = 1/R => R = 10^3 = 1000 ohm.
Now, from second circuit, i1 = Eeq / (R + req) = 3 / 1000 = 3 mA.
Equivalent voltage = (e1*r1 + e2*r2)/(r1 + r2) = (1*0.1 + 5*0.1)/(0.1 + 0.1) = 0.6 / 0.2 = 3 V
From first circuit, we have i = e/(R+r). Value of r is very less and can be neglected as compared to R.
Now, i = e/R => 10^-3 = 1/R => R = 10^3 = 1000 ohm.
Now, from second circuit, i1 = Eeq / (R + req) = 3 / 1000 = 3 mA.