Given:
The value of each resistance in parallel combination is,
[tex]15.0\text{ ohm}[/tex]
The value of the resistance in series with the parallel resistances is,
[tex]10.0\text{ ohm}[/tex]
The potential drop across the battery is,
[tex]45.0\text{ V}[/tex]
To find:
The voltage drop across each 15.0 ohm resistance
Explanation:
The value of the equivalent resistance of the three parallel resistances is,
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{15.0}+\frac{1}{15.0}+\frac{1}{15.0} \\ \frac{1}{R}=\frac{3}{15.0} \\ R=\frac{15.0}{3} \\ R=5.0\text{ ohm} \end{gathered}[/tex]
The net resistance of the circuit is,
[tex]\begin{gathered} R_{net}=10.0+5.0 \\ =15.0\text{ ohm} \end{gathered}[/tex]
The circuit diagram is like this:
The current in the circuit is,
[tex]\begin{gathered} I=\frac{V}{R_{net}} \\ =\frac{45.0}{15.0} \\ =3.0\text{ A} \end{gathered}[/tex]
The potential drop across 10.0 ohm is,
[tex]\begin{gathered} I\times10.0 \\ =3.0\times10.0 \\ =30.0\text{ V} \end{gathered}[/tex]
The potential drop across each of the 15.0 ohm resistance will be the same as the resistances are parallel.
So, the potential drop across each 15.0 ohm is,
[tex]\begin{gathered} 45.0-30.0 \\ =15.0\text{ V} \end{gathered}[/tex]
Hence, the required potential drop is 15.0 V.
