A battery of 6
6
V gives a current of 2
2
A when connected to a resistance of 2
2
ohm. What is the internal resistance, terminal PD and the lost voltage of the battery?
Total resistance, =voltagecurrent=62=3
R
=
voltage
current
=
6
2
=
3
ohm.
The external resistance, =2
R
e
=
2
ohm.
⇒
⇒
The internal resistance, =−=3−2=1
R
i
=
R
−
R
e
=
3
−
2
=
1
ohm.
⇒
⇒
The terminal PD = external voltage drop ==2×2=4
=
I
R
e
=
2
×
2
=
4
volt.
⇒
⇒
Lost voltage = 6
6
V - 4
4
V = 2
2
V.
