## Solved Examples of Electricity

**Example 1: **A TV set shoots out a beam of electrons. The beam current is 10mA.

**(a) **How many electrons strike the TV screen in each second ?

(b) How much charge strikes the screen in a minute?

**Solution: ** Beam current, I = 10 µA = 10 ×10^{–6}A

Time, t = 1 s

So,

(a) Charge flowing per second,

Q = I × t = 10 × 10^{–6}A × 1s = 10 × 10^{–6}C

We known,

Charge on an electron = 1.6 × 10^{–19}C

So, No. of electrons striking the TV screen per second =

$\frac{\text{10}\times {\text{10}}^{\text{-6}}C}{1.6\times \u200a{10}^{\u201319}C}$= 6.25 × 1014

(b) Charge striking the screen per min

= (6.25 × 10^{14} × 60) × 1.6 × 10^{–19}C

= 6.0 × 10^{–3}C

**Example 2: **A current of 10A exists in a conductor. Assuming that this current is entirely due to the flow of electrons (a) find the number of electrons crossing the area of cross section per second, (b) if such a current is maintained for one hour, find the net flow of charge.

**Sol. ** Current, I = 10 A

Charge flowing through the circuit

in one second, = 10 C (∴

$\frac{Ch\mathrm{arg}e}{Time}$= Current)

(a) We know, Charge on an electron

= 1.6 × 10^{–19}C

So, No. of electrons crossing per second

=

$\frac{10\u200aC}{1.6\times \u200a{10}^{\u201319}}$= 6.25 × 1019

(b) Net flow of charge in one hour

= Current × Time

= 10 A × 1 h

10 A × (1 × 60 × 60 s) = 36000 C

**Example 3: **A current of 5.0 A flows through a circuit for 15 min. Calculate the amount of electric charge that flows through the circuit during this time.

**Solution. ** Given : Current, I = 5.0 A

Time, t = 15 min. = 15 × 60 s = 900 s

Then, Charge that flows through the circuit,

Q = Current × Time

= 5.0A × 900 s

= 4500 A.s = 4500 C

**Example 4: **A piece of wire is redrawn by pulling it until its length is doubled. Compare the new resistance with the original value.

**Sol. **Volume of the material of wire remains same. So, when length is doubled, its area of cross-section will get halved. So, if *l *and a are the original length and area of cross-section of wire,

Original value of the resistance, R =

$\rho \times \frac{\ell}{a}$and,

New value of the resistance,

R’ =

$\rho \times \frac{2\ell}{a/2}=\rho \frac{\ell}{a}\times 4=4R$**Example 5: **Calculate the resistance of 100 m long copper wire. The diameter of the wire is 1 mm.

**Solution: ** Using the relationship,

R =

We have, r = mm = 0.5 × 10^{–3} m

R =

R = 2.04 ohm

**Example 6:** If four resistances each of values 1 ohm are connected in series. Calculate equivalent resistance.

** ****Solution: ****In series, **

R_{1} = R_{2} = R_{3} = R_{2} = 1 ohm

**putting values, we get, **

R_{s} = 1 + 1 + 1 + 1 = 4

** ****Example 7: **Suppose a 6-volt battery is connected across a lamp whose resistance is 20 ohm the current in the circuit is 0.25 A, calculate the value of the resistance from the resistor which must be used.

**Solution: ** Lamp resistance, R = 20 ohm

Extra resistance from resistor, R = ?

(to be calculated)

For R and R’ in series,

Total circuit resistance, R_{s} = R + R’

From relation, (Ohm’s law) R_{s} =

Putting values, we get, R_{s} =

= 24 ohm

But R_{s} = R + R’

Hence R’ = R_{s} – R

= 24 – 20 = 4 ohm

Extra resistance from resistor,

** R’ = 4 ohm.**

**Example 8:** A resistance of 6 ohms is connected in series with another resistance of 4 ohms. A potential difference of 20 volts is applied across the combination. Calculate the current through the circuit and potential difference across the 6 ohm resistance.

**Solution: **For better understanding we must drawn a proper circuit diagram. It is shown in fig.

We use proper symbols for electrical components.Resistances are shown connected in series, with 20 V battery across its positive and negative terminals. Direction of current flow is also shows from positive terminal of the battery towards its negative terminal.

Potential difference, V = 20 V

Potential difference across 6 Ω,

V_{1} = ? (to be calculated)

Total circuit resistance = 10 Ω

From Ohm’s law, R_{s} =

Circuit current, **I = 2 ampere or (2A) **

Putting values, we get, V_{1} = 2 × 6 = 12 volts

Potential difference across 6 Ω resistance = 12 V

** ****Example 9:** Two resistances are connected in series as shown in the diagram.

(i) What is the current through the 5 ohm resistance ?

(ii) What is the current through R ?

(iii) What is the value of R ?

(iv) What is the value of V ?

** **

**Solution: ** First resistance, R_{1} = 5 Ω

(i) Current through 5 ohm resistance, I = ?

(ii) Current through R, I = ?

(iii) Value of second resistance, R = ?

(iv) Potential difference applied by the battery,

V = ?

(i) From Ohm’s law, R =

We have, I =

I == 2 ampere

Current through 5 Ω resistance = 2 ampere (2A).

(ii) Since R is in series with 5 Ω, same current will flow through it,

Current through R = 2 A.

(iii) From Ohm’s law, R =

R_{2} =

R_{2} == 3 ohms

Resistance R has value = 3 ohms.

(iv) From relation, V = V_{1} + V_{2}

V = 10 + 6 = 16 volts

V = 16 volts

**Example 10: **Resistors R_{1}, R_{2} and R_{3} having values 5Ω , 10Ω, and 30Ω respectively are connected in parallel across a battery of 12 volt. Calculate (a) the current through each resistor (b) the total current in the circuit and (c) the total circuit resistance.

**Solution: **Here,

R_{1} = 5W, R_{2} = 10W, R_{3} = 30 W, V = 12 V

(a) I_{1} = ? I_{2} = ? I_{3} = ?

(b) I = I_{1} + I_{2} + I_{3} = ?

(c) R_{p} = ?

(a) From relation, (Ohm’s law), R =

I =

Putting values, we get, I_{1} = = =2.4 A

I_{2} = = = 1.2 A

I_{3} = 0.4 A

(b) Total current, I = I_{1} + I_{2} + I_{3}

I = 2.4 + 1.2 + 0.4 = 4 A

(c) From relation =

R_{p} = 3 ohm.

**Example 11: **Resistors R_{1} = 10 ohms, R_{2} = 40 ohms, R_{3} = 30 ohms, R_{4} = 20 ohms, R_{5} = 60 ohms and a 12 volt battery is connected as shown. Calculate :

(a) the total resistance and

(b) the total current flowing in the circuit.

**Solution: **The situation is shown in (figure).

For R_{1} and R_{2} in parallel

or = 8 ohm

For R_{3}, R_{4} and R_{5} is parallel

or = 10 ohm.

(a) For and in series.

Total resistance, R = +

Putting values, we get, R = 8 + 10 = 18

**Total resistance, R = 18 ohms. Ans.**

(b) From relation, (Ohm’s law) R =

We have, I =

Putting values, we get, I = = = 0.67

**Total current, I = 0.67 A. Ans**

**Example 12: **In the circuit diagram given below. Find

(i) total resistance of the circuit

(ii) total current flowing in the circuit

(iii) potential difference across R_{1}

**Solution:** **(i) For total resistance **

8 W and 12 Ω are connected in parallel.

Their equivalent resistance comes in series with 7.2 Ω resistance as shown in fig.

With 7.2 Ω and 4.8 Ω in series

R_{s} = 7.2 + 4.8 = 12 Ω

Total circuit resistance = 12 ohms.

**(ii) For total current **

Total circuit resistance, R = 12 ohm

Potential difference applied, V = 6 V

I = ?

From Ohm’s law R =

I =

I = = 0.5

Total circuit current = **0.5 A Ans. **

**(iii) For potential difference across R**_{1 }

R =

V = IR

V_{1} = IR_{1}

V_{1} = 0.5 × 7.2

= 3.6 V

Potential difference across, V_{1} = **3.6 V. Ans**

**Example 13: **Three resistances are connected as shown in diagram through the resistance 5 ohms, a current of 1 ampere is flowing :

(i) What is the current through the other two resistors?

(ii) What is the potential difference (p.d.) across AB and across AC ?

(iii) What is the total resistance.

**Solution:**

**(i)** **For current in parallel resistors**

For same potential difference across two parallel resistors,

V = I_{1}R_{1} = I_{2}R_{2} i.e.

**Current divides itself in inverse ration of the resistances. **

Also total current, I = I_{1} + I_{2}

Also, I_{1} + I_{2} = 1 amp.

**I**_{1 }**= 0.6A, I**_{2 }**= 0.4 A. Ans.**

Current is **0.6 A** through 10 Ω

**(ii) For p.d. across AB**

From Ohm’s law, R = , V = IR

V = 1 × 5 = 5V

P.D. across AB = **5 V. Ans **

For parallel combination of 10 Ω and 15 Ω P.D. across BC, V = I_{1}R_{1} = 0.6 × 10 = 6 V

P.D. across AC = P.D. across AB + P.D. across BC.

= 5 + 6 = 11 V

**(iii) For total circuit resistance **

For 10 Ω and 15 Ω in parallel

R_{p} = = 6 Ω

Total resistance = 5 + 6 = 11 W

Total circuit resistance **= 11 Ω****. ****Ans**

**Example 14: **In the diagram shown below (Fig.), the cells and the ammeter both have negligible resistance. The resistors are identical. With the switch K open, the ammeter reads 0.6 A. What will be the ammeter reading when the switch is closed ?

**Solution: **Let the cell have potential difference V and each resistor have resistance R

**With key open**

Potential difference, = V

Circuit resistance of two parallel resistors,

Circuit current, I_{1} = 0.6 A

**With key closed **

Potential difference = V

Circuit resistance of three parallel resistors,

Circuit current, I_{2} = ?

For same potential difference V

V = I_{1} = I_{2}

I_{2} =

I_{2} = 0.6 × = 0.9

Circuit current with closed key = **0.9 A**.

**Example 15: **In the circuit diagram.

Find (i) total resistance

(ii) current shown by the ammeter A.

**Solution: **3 Ω and 2 Ω in series become 5 Ω. Equivalent circuit is shown in fig.

**(i) For total resistance **

R_{1} = R_{2} = 5 Ω are in parallel.

R_{p} = = = 2.5 Ω

Circuit resistance **= 2.5 ohm**

**(ii) For circuit current **

Potential difference, V = 4 V

Circuit resistance R_{p} = 2.5 Ω

Circuit current, I = ? (to be calculated)

From Ohm’s law, R =

I =

I = = 1.6 A

Circuit current = **1.6 A **

Ammeter reads circuit current **1.6 A**

**Example 16: **For the circuit shown in the following diagram what is the value of

(i) current through 6 Ω resistor

(ii) potential difference (p.d.) across 12 Ω.

**Solution:** **(i) For current through 6 Ω**** **** **

Current from 4 V battery flows through first parallel branch having 6 Ω and 3 Ω in series.

Current in this branch

I = = = **0.44 A **

**(ii) For p.d. across 12 Ω**

Current through second parallel branch

I = = A

P.D. across 12 W, V = × 12 = 3.2 V.