
The potential difference across the resistance \(R\) is \(6\,\text{V}\).
Using Ohm’s law,
\[ i=\frac{V}{R} \]
From the circuit,
\[ i=\frac{10-6}{4} \]
\[ i=\frac{4}{4}=1\,\text{A} \]
Again using Ohm’s law for resistance \(R\),
\[ V=iR \]
Substituting the values,
\[ 6=1\times R \]
Therefore,
\[ R=6\,\Omega \]
Final Answer:
\[ \boxed{6\,\Omega} \]
Figure shows a part of an electric circuit. The potentials at points \( a, b, \text{and} \, c \) are \( 30 \, \text{V}, 12 \, \text{V}, \, \text{and} \, 2 \, \text{V} \), respectively. The current through the \( 20 \, \Omega \) resistor will be:

