Given:
The resistances R1 = 238 ohm, R2 = 125 ohm, and R3 =311 ohm are connected in series across the potential difference.
To find
(a) Equivalent resistance of the circuit.
(b) Current through R1.
(c) Voltage drop across R2.
Explanation:
(a) The equivalent resistance in the circuit is
[tex]\begin{gathered} R_{eq}=R1+R2+R3 \\ =\text{ 674 ohm} \end{gathered}[/tex]In a series circuit, the current through the circuit will be same.
(b)The current through the resistance R1 will be
[tex]\begin{gathered} I=\frac{V}{R_{eq}} \\ =\frac{99.3}{674} \\ =0.147\text{ A} \end{gathered}[/tex](c) The voltage drop across R2 will be
[tex]\begin{gathered} V=IR2 \\ =0.147\times125 \\ =18.375\text{ V} \end{gathered}[/tex]Final Answer
(a) Equivalent resistance of the circuit is 674 ohm.
(b) Current through R1 is 0.147 A.
(c) Voltage drop across R2 is 18.375 V.