A 20V battery is connected to a complex circuit with three branches. The first branch has a 10Ω light bulb and a switch in series. The second branch has a 10Ω and 20Ω light bulb in series. The third branch has only the battery. Solve the circuit.

Explanation:

voltage of battery, V = 20 v

first branch: a 10 ohm bulb and a switch

second branch: a 10 ohm bulb in series with a 20 ohm bulb

Total resistance of the second branch as they are connected in series

Rs = R1 + R2

Rs = 10 + 20 = 30 ohm

Now they are connected in parallel, so the effective resistance of the circuit is R.

R = 10 x 30 / (10 + 30) = 300 / 40 = 7.5 ohm

total current in the circuit, i = V / R = 20 / 7.5 = 2.67 Amphere

Voltage across first branch , V1 = V = 20 V

Current in first branch, i1 = V / R' = 20 / 10 = 2 A

Voltage across second branch, V2 = V = 20 V

current  in second branch, i2 = i - i1 = 2.667 - 2= 0.667 A

the answer for apex is the lowest

f is prop to m1xm2 double m1 double force