Answer to Question #241282 in Electric Circuits for shalline

Question

b)   Three resistors, two of 4K8 both in parallel and a third one, are connected in series. A fourth resistor of 4M5 and fifth of twice the value of the third resistor together in series are connected in parallel with this circuit. The equivalent resistance of all the five resistors is given by 3K5.

                       i.        Calculate the value of the third and fifth resistors.

                      ii.        If a battery with e.m.f. of 20V is connected to this final circuit, calculate the current through the fifth resistor. Repeat for capacitors.            

Answer 1

Explanations & Calculations

• First of all, you need to calculate the equivalent resistance of the entire circuit. Once it is found comparing it with the given value of 35K, the unknown resistance can be calculated.
• Those two 48 K are in parallel & their equivalent is $\small 48/2=24K$ and this is in series with the third one, say $\small R_3$. And their equivalent is $\small (24K+R_3)$.
• On the other branch, those 2 are in series connection & their equivalent is $\small (45000K+R_5)$ having taken the other unknown resistance to be $\small R_5$.
• Now, those 2 equivalents are connected parallelly & their equivalent or the equivalent of the entire circuit is

$\qquad\qquad \begin{aligned} \small \frac{1}{R_{E}}&=\small \frac{1}{(24+R_3)}+\frac{1}{(45000+R_5)}\\ \small \frac{1}{35}&=\small \frac{1}{(24+R_3)}+\frac{1}{(45000+2R_3)}\\ \small R_3&=\small 1.46K\\ \therefore\,\small R_5&=\small 2R_3=2.92K \end{aligned}$

• In finding the unknown resistance, all the values were taken to a common unit: $\small k\Omega\,or\,K$.

• Now you know the total resistance of the circuit. Then using $\small V =iR$ you can calculate the total current ($\small I$ ) flown in the circuit.
• Once the total current is known, consider the intermediate situation where we had the equivalent resistances of the two branches.
• There the total current is split into the two branches according to their resistance. Then the current flowing in the branch with the resistance $\small (45000K+R_5)$ is the current that flows through the fifth resistor.
• Then it will be,

$\qquad\qquad \begin{aligned} \small i&=\small I\times\frac{(24+R_3)}{(24+R_3)+(45000+R_5)} \end{aligned}$

• Give it a try.