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Question
Figure 6 below shows an electric circuit.

Apply Kirchhoff’s laws to calculate:
- current ‘I’ flowing through the 2Ω resistor.
- emf of the cell ‘E’.
Numerical
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Solution
a. Applying kirchhoff’s junction rule at the point Q,
I + 3 = 4
⇒ I = 4 − 3
⇒ I = 1 A
b. Applying Kirchhoff’s voltage rule in the loop SQRS, we get,
E = (2 × 1 + 4 × 1)
= 6 V
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