Current and Power Calculations in Series and Parallel Circuits

Question 1

there is a circuit which has a 6V cell with 2 and 3 ohm resistors in parallel. what is the current in each resistor

Answer 1

• I = V / R
• 6 / 2 = 3A
• 6 / 3 = 2A

Question 2

what is the power dissipated in each resistor

Answer 2

• you only have I and R so you cant use P = IV
• instead you use P = I^2R
• 3^2 x 2 = 18W
• 2^2 X 3 = 12W

Question 3

what is the power developed by the battery

Answer 3

• P = IV
• current is 3A + 2A = 5A
• 6 x 5 = 30W
• or 12 + 18 = 30W

Question 4

what do you notice about the answers to the last two questions

Answer 4

• the sum of power dissipated by the resistors equals the power developed by the cell
• this makes sense due to the conservation of energy
• all the pd (or power generated by the cell) in the circuit has to be used up

Question 5

the cell has an internal resistance of 0.8 ohms. what is the new total resistance of the circuit

Answer 5

• the two resistors in parallel:
• 1 /
• 1 / (1/2 + 1/3) = 1.2
• 1.2 + 0.8 in the cell = 2 ohms

Question 6

what is the total current in the circuit

Answer 6

• I = total V / total R

- 6 / 2 = 3A

Question 7

how would you calculate the potential difference across the two parallel resistors or just around the circuit in general

Answer 7

• V = e - Ir
• V = 6 - (3 x 0.8)
• 6 - 2.4 = 3.6
• V = 3.6

Question 8

what would the current in each resistor now be

Answer 8

• I = V / R
• 3.6 / 2 =1.8A
• 3.6 / 3 = 1.2A
• this naturally = 3A as previously calculated

Question 9

what is then the power dissipated by each resistor

Answer 9

• P = I^2R
• 1.8^2 x 2 = 6.5W
• 1.2^2 x 3 = 4.3W

Question 10

what is the power developed by the battery

Answer 10

• P = Ie (IV)

- 6 x 3 = 18W

Question 11

why is the power developed by the battery not equal to the power dissipated by the resistors

Answer 11

• because some power is wasted in the battery

- as it is used to overcome the internal resistance

Question 12

what is then the actual value of the net power developed by the battery that is useful

Answer 12

• 3.6 x 3 = 10.8W
• or you could calculate the power used by the internal resistance
• 6 - 3.6 (V in the circuit) = 2.4
• P = IV so 2.4 x 3 = 7.2W
• 18 - 7.2 = 10.8W
• or just add the power dissipated by the resistors
• 6.5 + 4.3 = 10.8W