Higher

Question
Circuit diagram
  1. Calculate the total resistance of the circuit.
  2. Calculate the current flowing in the {3}\Omega resistor.
  3. Calculate the voltage across the {8}\Omega resistor.
  4. Calculate the current flowing in the {6}\Omega resistor.

1. First calculate the parallel resistance:

\frac{1}{R}=\frac{1}{R}_{1} +\frac{1}{R}_{2}

R1 = {8}\Omega

R2 = {6}\Omega

\frac{1}{R}=\frac{1}{8} +\frac{1}{6}

\frac{1}{R}=\frac{7}{24}

R = \frac{24}{7}

R = {3.43}\Omega

This is in series with the {3}\Omega. The total resistance R is given by:

R = R1 + R2

R = {3}\Omega + {3.43}\Omega

R = {6.43}\Omega

The total resistance of the circuit is {6.43}\Omega.

2. The circuit above has now been simplified to:

Simplified circuit

Hence, the voltage across the {6.43}\Omega resistor is 12 V.

I = \frac{V}{R}

V = 12 V

R = {6.43}\Omega

I = \frac{12 V}{{6.43}\Omega}

I = 1.87 A

The current is the same in all parts of a series circuit and so this is the current through the {3}\Omega resistor too.

The current flowing in the {3}\Omega is 1.87 A.

3. The current through the {8}\Omega resistor is not known so we cannot use V = IR to calculate the voltage across it directly.

However, we do know the current through the {3}\Omega resistor, so we can calculate the voltage across it.

V = IR

I = 1.87 A

R = {3}\Omega

V = 1.87 A x {3}\Omega

V = 5.61 V

The total voltage = 12 V

Hence, the voltage across the two parallel resistors = 12 V – 5.61 V = 6.39 V.

The voltage across each parallel resistor is the same.

Hence, the voltage across the {8}\Omega = 6.39 V.

The voltage across the 8Ω resistor is 6.39 V.

4. The voltage across each parallel resistor is the same.

Hence, the voltage across the {6}\Omega resistor = 6.39 V.

I = \frac{V}{R}

V = 6.39 V

R = {6}\Omega

I = \frac{6.39 V}{{6}\Omega}

I = 1.07 A

The current through the {6}\Omega is 1.07 A.