Three small lamps, R1 = 4.2 , R2 = 2.8 , and R3 = 1.6 are connected to a 9.0 V battery, as shown below.

Question

(a) What is the equivalent resistance of the circuit?

(b) What is the current in the battery?

(c) What is the current in each bulb?

(d) What is the potential difference across each bulb?

Answer 1

(a) First, get the resistance of R2 and R3 in parallel. R = R2xR3 / R2 + R3 = 2.8x1.6 /(2.8+1.6) = 1.02 ohms

Next add R1 to R to get 4.2 + 1.02 = 5.22 ohms which is the equivalent resistance of the entire circuit.

(b) The total current is I = E/R = 9.0 V / 5.22 ohms = 1.72 amps.

(c) The current in R1 is the total current = 1.72 amps. The voltage drop across the parallel branch is the total current times it's equivalent resistance or 1.72 x 1.02 = 1.75 volts. The current through R2 is therefore 1.75 V /2.8 = 0.63 amps. The current through R3 is 1.75 V / 1.6 = 1.09 amps.

(d) The potential difference across R1 is 1.72 amps x 4.2 ohms = 7.22 volts.

Note that the sum of the potential drops across R1 and the equivalent parallel resistance of R2 + R3 is 7.22 + 1.75 = 8.97 volts which is very close to the battery voltage. The difference between the given battery voltage and the calculated voltage is due to rounding.

Please check my math.