In the circuit of Fig. 2.67, a decrease in R3 leads to a decrease of:(a) current through R...

Find the current flow through resistor \(R_{3}\) using current division rule.

$$ \begin{array}{l} i_{3}=i\left(\frac{R_{2}}{R_{2}+R_{3}}\right) \\ \text { Assume } i\left(\frac{R_{2}}{R_{2}+R_{3}}\right)=i_{o} \end{array} $$

Consider resistor \(R_{3}\) is decreased to \(\frac{R_{3}}{2}\).

Now current \(i_{3}\) becomes:

$$ \begin{aligned} i_{3(\mathrm{new})} &=i\left(\frac{R_{2}}{R_{2}+\frac{R_{3}}{2}}\right) \\ &=2 i\left(\frac{R_{2}}{2 R_{2}+R_{3}}\right) \end{aligned} $$

Thus, current \(i_{3(\mathrm{new})}\) is increased compared to current \(i_{o}\).

So, option (a) is wrong option.

Find the voltage across resistor \(R_{3}\).

$$ V_{3}=i_{3} R_{3} $$

Assume the voltage across resistor \(R_{3}\) as \(V_{3}=v_{o}\)

Consider resistor \(R_{3}\) is decreased to \(\frac{R_{3}}{2}\).

$$ \begin{aligned} V_{3(\mathrm{new})} &=i_{3} \frac{R_{3}}{2} \\ &=\frac{V_{3}}{2} \end{aligned} $$

Thus, voltage \(V_{3(\mathrm{ncw})}\) is decreased compared to voltage \(v_{o} .\) So, option (b) is correct option.

Find the current flow through the resistor \(R_{1}\).

$$ i_{1}=\frac{V_{s}}{R_{e q}} \ldots \ldots .(1) $$

Here,

$$ \begin{aligned} R_{e q} &=R_{1}+R_{2} \| R_{3} \\ &=R_{1}+\frac{R_{2} R_{3}}{R_{2}+R_{3}} \end{aligned} $$

Substitute \(R_{1}+\frac{R_{2} R_{3}}{R_{2}+R_{3}}\) for \(R_{c q}\) in equation (1).

$$ \begin{aligned} i_{1} &=\frac{V_{s}}{R_{1}+\frac{R_{2} R_{3}}{R_{2}+R_{3}}} \\ &=\frac{V_{s}}{R_{1}+\frac{R_{2} R_{3}}{R_{2}+R_{3}}} \end{aligned} $$

Find the voltage across resistor \(R_{1}\).

$$ \begin{aligned} V_{1} &=R_{1} i_{1} \\ &=R_{1}\left(\frac{V_{s}}{R_{1}+\frac{R_{2} R_{3}}{R_{2}+R_{3}}}\right) \end{aligned} $$

Consider resistor \(R_{3}\) is decreased to \(\frac{R_{3}}{2}\).

$$ \begin{aligned} V_{1}(\text { new }) &=R_{1} i_{1} \\ &=R_{1}\left(\frac{V_{s}}{R_{1}+\frac{R_{2} \frac{R_{3}}{2}}{R_{2}+\frac{R_{3}}{2}}}\right) \\ &=\frac{V_{s} R_{1}}{R_{1}+\frac{R_{2} R_{3}}{2 R_{2}+R_{3}}} \end{aligned} $$

From the expression it is clear that denominator reduced which leads to overall voltage

\(V_{1}\)

increases.

Thus, the voltage across resistor \(R_{1}\) is increases by decreasing resistor \(R_{3}\).

So, option (c) is wrong option.

Determine the power dissipated in \(R_{2}\).

$$ \begin{aligned} P_{R_{2}} &=i_{2} v_{2} & & \\ &=i_{2} v_{3} \text { since } v_{3}=v_{2} & \\ &=i_{2}\left(R_{3} i_{3}\right) & & \text { since } v_{3}=R_{3} i_{3} \\ &=\left(\frac{i_{1} R_{3}}{R_{2}+R_{3}}\right) i_{3} R_{3} & \text { since } i_{2}=\frac{i_{1} R_{3}}{R_{2}+R_{3}} \\ &=\left(\frac{V_{s}}{R_{1}+\frac{R_{2} R_{3}}{R_{2}+R_{3}}}\right)\left(\frac{R_{3}^{2} i_{3}}{R_{2}+R_{3}}\right) & \text { since } i_{1}=\frac{V_{s}}{R_{1}+\frac{R_{2} R_{3}}{R_{2}+R_{3}}} \end{aligned} $$

Consider resistor \(R_{3}\) is decreased to \(\frac{R_{3}}{2}\)

$$ \begin{aligned} P_{R_{2}(\mathrm{ew})}=& \frac{V_{s}}{R_{1}+\frac{R_{2} \frac{R_{3}}{2}}{R_{2}+\frac{R_{3}}{2}}}\left(\frac{\frac{R_{3}}{2}}{R_{2}+\frac{R_{3}}{2}}\right) i_{3} \frac{R_{3}}{2} \\ =& \frac{V_{s}}{R_{1}+\frac{R_{2} R_{3}}{2 R_{2}+R_{3}}}\left(\frac{\frac{R_{3}}{2}}{R_{2}+\frac{R_{3}}{2}}\right) i_{3} \frac{R_{3}}{2} \end{aligned} $$

$$ =\left(\frac{V_{s}}{R_{1}+\frac{R_{2} R_{3}}{2 R_{2}+R_{3}}}\right)\left(\frac{i_{3} R_{3}^{2}}{4\left(2 R_{2}+R_{3}\right)}\right) $$

It is clear that \(P_{R_{2}(\mathrm{new})}\) is decreased compared to \(P_{R_{2}}\).So, option (d) is correct option.

Therefore, voltage across \(R_{3}\) and power dissipated in \(R_{2}\) decreases with decrease in the value of resistor \(R_{3}\).

So correct options are (b) and (d).