Simple Electrical Circuit Analysis

00:00 5 marks

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6. A simple electrical circuit consists of a variable resistor R connected to a cell of emf 6.0 V and internal resistance 1.5 Ω.nnnn(a) Define *electromotive force (emf)*. [1]nn(b) The variable resistor is set to 4.5 Ω. Calculate:n (i) the current in the circuit. [2]n (ii) the power dissipated in the internal resistance. [2]

nn Simple Circuit with Internal Resistancen A circuit diagram showing a cell with EMF 6.0V and internal resistance 1.5 Ohm in series with a variable resistor R.nn nn n n n nn n n n n n n εn = 6.0 Vn nn n n n n n rn = 1.5 Ωn n nn n n n nn n n n n n n n n n n Rn nn n n n n n nn

Part a

1 minutes 1 marks

Question

(a) Define *electromotive force (emf)*. [1]



Hint Hint

Hint

Think about the energy conversion that happens inside a battery or cell.

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Complete Answer

Emf is the work done per unit charge in moving charge completely around the circuit / energy converted from chemical (or other forms) to electrical per unit charge by the source.

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Part b(i)

2 minutes 2 marks

Question

(b) The variable resistor is set to 4.5 Ω. Calculate:n (i) the current in the circuit. [2]



Hint Hint

Hint

Use Ohm’s law for the entire circuit. Remember to include both the external and internal resistance.

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Complete Answer

1.0 A

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Part b(ii)

2 minutes 2 marks

Question

(ii) the power dissipated in the internal resistance. [2]



Hint Hint

Hint

Use the formula for power dissipated in a resistor (P = I²R) and the current calculated in part (b)(i). Apply it to the internal resistance.

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Complete Answer

1.5 W

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