Question

In the network shown below, the potential difference across CD is

1. 4 V
2. 6 V
3. 10 V
4. 5 V

Answer 1

Hint: Using the formula for computing the resistance in parallel, we will first compute the parallel resistance value. Then, using the formula for computing the resistance in series, we will compute the series resistance, that is, equivalent resistance value. Finally, we will subtract the total voltage by the voltage across 40 Ohm resistance using the product of current and resistance.

Formula used:
\[\begin{align}
  & \dfrac{1}{{{R}_{P}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+.... \\
 & {{R}_{S}}={{R}_{1}}+{{R}_{2}}+... \\
 & V=IR \\
\end{align}\]

Complete step-by-step solution:
Firstly, we will compute the parallel resistance value of 2 resistors. The formula for computing the same is given as follows.
\[\dfrac{1}{{{R}_{P}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}\]
Substitute the values in the above formula.
\[\begin{align}
  & \dfrac{1}{{{R}_{P}}}=\dfrac{1}{100}+\dfrac{1}{150} \\
 & \therefore {{R}_{P}}=75\Omega \\
\end{align}\]
 Now, we will compute the parallel resistance value of 2 resistors. The formula for computing the same is given as follows.
\[{{R}_{S}}={{R}_{1}}+{{R}_{2}}\]
Substitute the values in the above formula.
\[\begin{align}
  & {{R}_{S}}=75+40 \\
 & \therefore {{R}_{S}}=115\Omega \\
\end{align}\]
Therefore, the equivalent resistance of the circuit is, \[115\,\Omega \].
The current flowing through the circuit is given as the voltage by the resistance. The mathematical representation of the same is given as follows.
\[I=\dfrac{V}{R}\]
Substitute the values in the above formula.
\[\begin{align}
  & I=\dfrac{10}{115} \\
 & \therefore I=0.086A \\
\end{align}\]
Now we will compute the potential difference across the 40 Ohm resistor. The formula for computing the potential difference is the product of current and the resistance.
The mathematical representation of the same is given as follows.
\[V=IR\]
Substitute the values in the above formula.
\[\begin{align}
  & V=0.086\times 40 \\
 & \therefore V=3.44\,V \\
\end{align}\]
Finally, we will compute the potential difference across CD, so, we have,
\[\begin{align}
  & {{V}_{CD}}=V-{{V}_{40\Omega }} \\
 & \Rightarrow {{V}_{CD}}=10-3.44 \\
 & \therefore {{V}_{CD}}=6.56\,V \\
\end{align}\]
\[\therefore \] The potential difference across the CD is, 6.56.

Note: Instead of computing the potential difference across CD, subtracting the total voltage by the voltage across 40 Ohm resistance using the product of current and resistance becomes easier. The units of the parameters should be taken care of.