Electrical Circuit-Solved Numericals

1. A cell of emf 2v and internal resistance 1ohm are connected with two resistors of resistances  2 and 5ohms in parallel. Find the current in each resistor.

Given:

EMF of a cell (E) = 2V

Internal resistance (r) = 1ohm

External resistance (R1) = 2ohm

R2 = 5ohm.

Equivalent resistance (R) =R1R2R1+R2$\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_{1+}{\mathrm{R}}_2}$107$\frac{10}{7}$ ohm.

Now,

Current I = ER+r$\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}$ = 2107+1$\frac{2}{\frac{10}{7}+1}$

I = 1417$\frac{14}{17}$A.

Again, voltage through 2ohm = voltage through 5 ohm.

Or, 2I1 = 5I2

Or, I2 = 25$\frac{2}{5}$I1

Total current of circuit (I) = I1 + I2

Or, 1417=I1+25I1$\frac{14}{17}={\mathrm{I}}_1+\frac{2}{5}{\mathrm{I}}_1$

Or, 1417=75I1$\frac{14}{17}=\frac{7}{5}{\mathrm{I}}_1$

Or, I1 = 1017$\frac{10}{17}$ohm.

So, I2 = 25$\frac{2}{5}$ * I1

= 25$\frac{2}{5}$ * 1017$\frac{10}{17}$ ohm

= 417ohm$\frac{4}{17}\mathrm{o}\mathrm{h}\mathrm{m}$

So, the current in 2ohm and 5ohm is 1017$\frac{10}{17}$ A and 417$\frac{4}{17}$A respectively.

2. In the circuit as shown in the figure,calculate the current through each cell and through resistance R.

Soln:

EMF of battery (E1) = 6V

Internal resistance (r1) = 4ohm

E2 = 4V

r2 = 2hm

External resistance (R) = 1ohm.

We have

Total current in the circuit I= I1+I2

Applying KVL to the first mesh,

(I1+I2 )1+4 I1=6 ……………i

Applying KVL to the secondmesh,

(I1+I2 )1+2 I2=4………..ii

Solving we get

I1= 1A

I2 =1A

 I = 2A

∴ current through the resistance R = 2A

3. A battery of emf 6v and 0.1 internal resistance is joined in parallel with another battery of emf 10v and 1ohm internal resistance. The combination is used to send current through an external resistance of 12ohm . Calculate the current through each battery.

Soln:

EMF of battery (E1) = 6V

Internal resistance (r1) = 0.1ohm

E2 = 10V

r2 = 1ohm

External resistance (R) = 12ohm.$$

We have

Total current in the circuit I= I1+I2

Applying KVL to the first mesh,

(I1+I2 )12+0.1 I1=6 ……………i

Applying KVL to the secondmesh,

(I1+I2 )12 +1 I2=10………..ii

Solving we get

I1=6019$\frac{60}{19}$A

I2 =7019$\frac{70}{19}$A

 I = 2A

∴ current through the resistance R = 13019A$\frac{130}{19}\mathrm{A}$

4. Find the equivalent resistance between A and B in the following networks.

The above circuit can be drawn as:

The given setup forms a wheat stone’s bridge in balanced position, hence P and O are at same potential and no current flows in resistance 5ohm between P and O. Hence, this resistance may be ignored.

For resistance between A and B the resistance 2X 5 between Apo and 2 * 5 between BPO are in parallel.

So,

Or, 1RAB=110+110$\frac{1}{{\mathrm{R}}_{\mathrm{A}\mathrm{B}}}=\frac{1}{10}+\frac{1}{10}$

So, RAB = 5ohm.

5.A potentiometer wire of length 100cm has an internal resistance of 10ohm. It is connected in series with a resistance and cell of emf 2v and negligible internal resistance. A source of emf 10milliv is balanced by a length of 40cm of the potentiometer wire. Find the value of the resistance.

 Given:

Length of wire (l1) = 100cm = 1m

Resistance of wire (R1) = 10ohm

EMF of cell (E1) = 2V

Internal resistance = 0

Balancing length (l2) = 40cm = 0.4m

Balancing EMF (E2) = 10mV

= 0.01V.

External resistance (R) = ?

The current in Potentiometer circuit, I = 2R+10$\frac{2}{\mathrm{R}+10}$

Now,

PD across 40cm length of wire=10mV

PD across 100cm length of wire=1040∗100mV=25mv$\frac{10}{40}\ast 100\mathrm{m}\mathrm{V}=25\mathrm{m}\mathrm{v}$

PD across AB length of wire

VAB = I.RAB

Or, 251000$\frac{25}{1000}$ = 2R+10$\frac{2}{\mathrm{R}+10}$ * 10.

So, R = 790ohm.

6.P,Q,R and X are four coils of wire 2,2,2 and 3ohm rresistance respectively arranged to form a wheatstone bridge. Calculate the value of resistance with which the coil X must be shunted in order that the bridge may be balanced.

If P,Q,R and S are resistances of wheat stones bridge, then

In balanced position,

Or, PQ=RS$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}$

Here,P = 2ohm, R = 2ohm, S = 3ohm

Then,

Q = SR.P$\frac{\mathrm{S}}{\mathrm{R}}.\mathrm{P}$

= 3ohm.

i.e. In arm BC, the net resistance should be 20ohm, but it contains a combination of resistances 2 ohm and R’ in parallel so,

or, 13=12+1R′$\frac{1}{3}=\frac{1}{2}+\frac{1}{{\mathrm{R}}^{\prime }}$

or, 1R′=−16$\frac{1}{{\mathrm{R}}^{\prime }}=-\frac{1}{6}$.

So, Resistance can’t n be negative.

So, R’ = 6ohm.

7.In a potentiometer arrangement,a cell of 1.25V gives a balance point at 35cm length of the wire. If the cell is replaced by another cell and the balanced point shifs to 63cm what is the emf of the second cell?

 EMF of 1st cell (E1) = 1.25V

Length (l1) = 35cm = 0.35m

Balancing length of 2nd cell (l2) = 63cm

= 0.63m

E2= ?.

Or, E1E2=l1l2$\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}$

Or, 1.25E2=0.350.63$\frac{1.25}{{\mathrm{E}}_2}=\frac{0.35}{0.63}$

So, E2 = 2.25V.

8.The driver cell of potentiometer has an emf 2v and negligible internal resistanc . the potentiometer wire has a resistance 3ohm. Calculate th resistance needed in series with the wire if the pd of 5mv is required across the whole wire.

Given,

EMF of cell (E) = 2V

r = 0.

Resistance (R1) = 3 ohm.

Let R be the resistance to be added across the wire,

Then,

PD (V) = 5mV = 5 * 10-3V.

The current in potentiometer circuit (I) = ER1+R$\frac{\mathrm{E}}{{\mathrm{R}}_1+\mathrm{R}}$

Or, I = 23+R$\frac{2}{3+\mathrm{R}}$

Now,

PD across the whole wire

V = I.R1

Or, 5 * 10-3 = 23+R$\frac{2}{3+\mathrm{R}}$ * 3

Or, 0.015 + 5 * 10-3R = 6.

R = 1197 ohm.

9.The emf of a  battery A is balanced by a length of 75cm on a potentiometer wire.The standard cell of emf 1.02v is balanced by a length of 50cm. What is the emf of cell A?

Let EMF of battery A be E1.

Balancing length (l1) = 75cm = 0.75m

EMF of standard cell (E2) = 1.02V

Balancing length (l2) = 50cm = 0.5m

We have,

Or, E1E2=l1l2$\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}$

Or, E1 = E2 * l1l2$\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}$

= 1.02 * 0.750.5$\frac{0.75}{0.5}$

= 1.53V.