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

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 = \({3}\Omega + {3.43}\Omega\)

R = \({6.43}\Omega\)

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

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

I = \(\frac{V}{R}\)

R = \({6.43}\Omega\)

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

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.

R = \({3}\Omega\)

V = 1.87 A x \({3}\Omega\)

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

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

I = \(\frac{V}{R}\)

R = \({6}\Omega\)

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

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