In the circuit diagram given below, find:
(A) total resistance of the circuit,
(B) total current flowing in the circuit, and
(C) the potential difference across $R_{1}$

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Solution

When 8 and 12 are in parallel
and 7.2 is in series with them
Net Resistance $\frac{7.2}{10} + \frac{8 \times 12}{12 + 8}$
$= \frac{72}{10} + \frac{96}{20} = \frac{72}{10} + \frac{48}{10} = \frac{120}{10} = 12 Ω$
Total current $= \frac{V}{n e t R} = \frac{6}{12} = 0.5 Ω$
Voltage across $R_{1} = i R_{1} = 0.5 \times 7.2 = 3.6$

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In the circuit diagram given below, find:(A) total resistance of the circuit,(B) total current flowing in the circuit, and(C) the potential difference across R1

When 8 and 12 are in parallel and 7.2 is in series with them Net Resistance 7.210+8×1212+8=7210+9620=7210+4810=12010=12ΩTotal current =VnetR=612=0.5ΩVoltage across R1=iR1=0.5×7.2=3.6

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