A potential difference of 4 V is applied to two resistors of 6 $Ω$ and 2 $Ω$ connected in series. Calculate : (a) the combined resistance (b) the current flowing (c) the potential difference across the 6 $Ω$ resistor

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Solution

(a) combined resistance in series= $R_{1} + R_{2} = 6 + 2 = 8 Ω$

(b) $I = \frac{V}{R} = \frac{4}{8} = 0.5 A$

(C) p.d across 6 ohm= $I R = 6 \times 0.5 = 3 V$

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A potential difference of 4 V is applied to two resistors of 6Ω and 2 Ω connected in series. Calculate : (a) the combined resistance (b) the current flowing (c) the potential difference across the 6 Ω resistor

(a) combined resistance in series= R1+R2=6+2=8Ω(b)I=VR=48=0.5A(C) p.d across 6 ohm= IR=6×0.5=3V

A potential difference of 4 V is applied to two resistors of 6 $Ω$ and 2 $Ω$ connected in series. Calculate : (a) the combined resistance (b) the current flowing (c) the potential difference across the 6 $Ω$ resistor

(a) combined resistance in series= $R_{1} + R_{2} = 6 + 2 = 8 Ω$

(b) $I = \frac{V}{R} = \frac{4}{8} = 0.5 A$

(C) p.d across 6 ohm= $I R = 6 \times 0.5 = 3 V$