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^–6A

Time, t = 1 s

So,

(a) Charge flowing per second,

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

We known,

Charge on an electron = 1.6 × 10^–19C

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

\frac{10 \times 10^{-6} C}{1.6 \times 10^{- 19} C}

= 6.25 × 1014

(b) Charge striking the screen per min

= (6.25 × 10¹⁴ × 60) × 1.6 × 10^–19C

= 6.0 × 10^–3C

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{C h \arg e}{T i m e}

= Current)

(a) We know, Charge on an electron

= 1.6 × 10^–19C

So, No. of electrons crossing per second

\frac{10 C}{1.6 \times 10^{- 19}}

= 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 =

ρ \times \frac{ℓ}{a}

and,

New value of the resistance,

R’ =

ρ \times \frac{2 ℓ}{a / 2} = ρ \frac{ℓ}{a} \times 4 = 4 R

Example 5: Calculate the resistance of 100 m long copper wire. The diameter of the wire is 1 mm.

Solution: Using the relationship,

R = $\displaystyle \rho \times \frac{\ell }{a}=\rho \times \frac{\ell }{{\pi {{r}^{2}}}}$$

We have, r = $\frac{1}{2}$$ mm = 0.5 × 10^–3 m

R = $\frac{{1.6\times {{{10}}^{{-6}}}ohm.cm\,\times 100m}}{{3.141\times \,{{{(0.5\,\times \,{{{10}}^{{-3}}}m)}}^{2}}}}$$

R = 2.04 ohm

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

Solution: In series,

R₁ = R₂ = R₃ = R₂ = 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 = $\frac{V}{I}$$

Putting values, we get, R_s = $\frac{6}{{0.25}}$$

= 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,

V₁ = ? (to be calculated)

Total circuit resistance = 10 Ω

From Ohm’s law, R_s = $\frac{V}{{{{R}_{S}}}}$$

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

Putting values, we get, V₁ = 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, R₁ = 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 = $\frac{V}{I}$$

We have, I = $\displaystyle \frac{V}{R}=\frac{{{{V}_{1}}}}{{{{R}_{1}}}}$$

I = $\frac{{10}}{5}$$ = 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 = $\frac{V}{I}$$

R₂ = $\frac{{{{V}_{2}}}}{I}$$

R₂ = $\frac{6}{2}$$ = 3 ohms

Resistance R has value = 3 ohms.

(iv) From relation, V = V₁ + V₂

V = 10 + 6 = 16 volts

V = 16 volts

Example 10: Resistors R₁, R₂ and R₃ 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,

R₁ = 5W, R₂ = 10W, R₃ = 30 W, V = 12 V

(a) I₁ = ? I₂ = ? I₃ = ?

(b) I = I₁ + I₂ + I₃ = ?

(a) From relation, (Ohm’s law), R = $\frac{V}{I}$$

I = $\frac{V}{R}$$

Putting values, we get, I₁ = $\frac{V}{{{{R}_{1}}}}$$ = $\frac{{12}}{5}$$ =2.4 A

I₂ = $\frac{V}{{{{R}_{2}}}}$$ = $\frac{{12}}{{10}}$$ = 1.2 A

I₃ $\frac{V}{{{{R}_{3}}}}=\frac{{12}}{{30}}$$ = 0.4 A

(b) Total current, I = I₁ + I₂ + I₃

I = 2.4 + 1.2 + 0.4 = 4 A

$\frac{1}{{{{R}_{p}}}}=\frac{1}{5}+\frac{1}{{10}}+\frac{1}{{30}}=\frac{{6+3+1}}{{30}}=\frac{{10}}{{30}}$$

R_p = 3 ohm.

Example 11: Resistors R₁ = 10 ohms, R₂ = 40 ohms, R₃ = 30 ohms, R₄ = 20 ohms, R₅ = 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 R₁ and R₂ in parallel

$\frac{1}{{{{R}_{{{{p}_{1}}}}}}}=\frac{1}{{{{R}_{1}}}}+\frac{1}{{{{R}_{2}}}}=\frac{1}{{10}}+\frac{1}{{40}}=\frac{{4+1}}{{40}}=\frac{5}{{40}}=\frac{1}{8}$$

or ${{R}_{{{{P}_{1}}}}}$$ = 8 ohm

For R₃, R₄ and R₅ is parallel

$\displaystyle \frac{1}{{{{R}_{{{{p}_{2}}}}}}}=\frac{1}{{{{R}_{3}}}}+\frac{1}{{{{R}_{4}}}}+\frac{1}{{{{R}_{5}}}}$

$=\frac{1}{{30}}+\frac{1}{{20}}+\frac{1}{{60}}=\frac{{2+3+1}}{{60}}=\frac{6}{{60}}=\frac{1}{{10}}$$

or ${{R}_{{{{P}_{2}}}}}$$ = 10 ohm.

(a) For and ${{R}_{{{{P}_{2}}}}}$$ 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 = $\frac{V}{I}$$

We have, I = $\frac{V}{R}$$

Putting values, we get, I = $\frac{{12}}{{18}}$$ = $\frac{2}{3}$$ = 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 R₁

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 = $\frac{V}{I}$$

I = $\frac{V}{R}$$

I = $\frac{6}{{12}}$$ = 0.5

Total circuit current = 0.5 A Ans.

(iii) For potential difference across R₁

R = $\frac{V}{I}$$

V = IR

V₁ = IR₁

V₁ = 0.5 × 7.2

= 3.6 V

Potential difference across, V₁ = 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 = I₁R₁ = I₂R₂ i.e. $\frac{{{{I}_{1}}}}{{{{I}_{2}}}}=\frac{{{{R}_{2}}}}{{{{R}_{1}}}}$$

Current divides itself in inverse ration of the resistances.

Also total current, I = I₁ + I₂

$\displaystyle \frac{{{{I}_{1}}}}{{{{I}_{2}}}}=\frac{{{{R}_{2}}}}{{{{R}_{1}}}}=\frac{{15}}{{10}}=\frac{3}{2}$$

Also, I₁ + I₂ = 1 amp.

I₁= 0.6A, I₂= 0.4 A. Ans.

Current is 0.6 A through 10 Ω

(ii) For p.d. across AB

From Ohm’s law, R = $\frac{V}{I}$$ , 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 = I₁R₁ = 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 = $\frac{{10\times 15}}{{10+15}}=\frac{{150}}{{25}}$$ = 6 Ω

Total resistance = 5 + 6 = 11 W

Total circuit resistance = 11 Ω. Ans

$\left[ {Also\,R\,=\,\frac{V}{I}\,=\,\frac{{11}}{1}\,=\,11\Omega } \right]$$

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