Question

In the circuit shown in figure

A. power supplied by the circuit is 200 W.
B. current flowing in the circuit is 5A
C. potential difference across `4Omega` resistance is equal to the potential difference across `6Omega` resistance
D. current in wire AB is zero.

Answer 1

Correct Answer - A::C
a., c.
Resistances `4 and 6 Omega` are short-circuited. Therefore, no
current will flow through these two resistances. Current
passing through the battery is `I = (20//2) = 10A`. This is also
the current passing in wire AB from B to A. Power supplied by
the battery is
`P = EI =(20)(10) = 200W`
Potential difference across the `4 Omega` resistance = potential
difference across the `6Omega` resistance = 0