Question

The following figure shows a circuit that contains two batteries and two resistors. Use Kirchoff’s Loop Rule to determine the current in circuit ?
Determine the circuit:

Answer 1

Hint: Kirchhoff’s loop rule states that sum of all the electric potential difference around a loop is zero i.e. \[\Delta V{\text{ }} = {\text{ }}0\] (written mathematically). This law goes in accordance with conservation of energy theory. And \[V = IR\]

Step by step complete answer-

Since it is a numerical problem, we will solve it by keeping the concept and rules in mind-
In case of \[12\Omega \] resistor current flows in the same direction of resistor and in case of \[6V\] battery the current is flowing from + to – terminal hence voltage is taken as positive.
In case of 8 Ω resistor current again flows in same direction as that of a resistor and in case of \[24V\] battery the current is moving from – to + terminal and hence voltage is taken as negative.
Therefore,
\[12 \times I + 6 + 8 \times I - 24 = 0\]
\[ \Rightarrow 20 \times I - 18 = 0\]
\[ \Rightarrow 20{\text{ }}I = 18\]
\[ \Rightarrow I = 18 \div 20\]
\[I = 0.9{\text{ }}A\]

Note:
On applying Kirchhoff’s rule and formula we obtain the answer as \[0.9A\] .
Rules-
If we cross a resistor which is in same direction as that of current flow, the sign for voltage is taken as +
In case of battery , the sign for voltage is taken as + if we go from positive terminal to negative terminal.