Class 10 - Unit # 15 : Current Electricity - Solved Numericals

1. When the current in a pocket calculator is $0.0002\text{ A}$, how much charge flows every minute?

Data:

• Electric current ($I$) = $0.0002\text{ A} = 2 \times 10^{-4}\text{ A}$
• Time interval ($t$) = $1\text{ minute} = 60\text{ seconds}$
• Charge ($Q$) = ?

Solution:

According to the fundamental definition of electric current:

$$I = \frac{Q}{t} \implies Q = I \cdot t$$ $$Q = (0.0002\text{ A}) \times 60\text{ s}$$ $$Q = 0.012\text{ C}$$

Converting the charge into milli-Coulombs ($\text{mC}$):

$$Q_{\text{mC}} = 0.012 \times 1000 = 12\text{ mC}$$

RESULT: The amount of charge that flows through the calculator is $12\text{ mC}$ (or $0.012\text{ C}$).

2. Calculate the amount of current that an electric heater uses to heat a room in $5\text{ minutes}$ if the charge transferred is $2100\text{ C}$.

Data:

• Total charge ($Q$) = $2100\text{ C}$
• Time ($t$) = $5\text{ minutes} = 5 \times 60\text{ s} = 300\text{ seconds}$
• Electric current ($I$) = ?

Solution:

Using the electric current formula:

$$I = \frac{Q}{t}$$ $$I = \frac{2100\text{ C}}{300\text{ s}}$$ $$I = 7\text{ A}$$

RESULT: The amount of current used by the electric heater is $7\text{ A}$.

3. A potential difference of $90\text{ V}$ exists between two points. The amount of work done when an unknown charge is moved between the points is $450\text{ J}$. Determine the charge amount.

Data:

• Potential difference ($V$) = $90\text{ V}$
• Work done ($W$) = $450\text{ J}$
• Charge ($Q$) = ?

Solution:

Using the relation between work, charge, and electric potential:

$$V = \frac{W}{Q} \implies Q = \frac{W}{V}$$ $$Q = \frac{450\text{ J}}{90\text{ V}}$$ $$Q = 5\text{ C}$$

RESULT: The amount of charge moved between the points is $5\text{ C}$.

4. Calculate the potential difference between two points A and B if it takes $9 \times 10^{-4}\text{ J}$ of external work to move a charge of $+9\text{ }\mu\text{C}$ from A to B.

Data:

• Work done ($W$) = $9 \times 10^{-4}\text{ J}$
• Charge ($Q$) = $+9\text{ }\mu\text{C} = 9 \times 10^{-6}\text{ C}$
• Potential difference ($V_{AB}$) = ?

Solution:

Using the definition of electric potential difference:

$$V_{AB} = \frac{W}{Q}$$ $$V_{AB} = \frac{9 \times 10^{-4}\text{ J}}{9 \times 10^{-6}\text{ C}}$$ $$V_{AB} = 10^{-4 - (-6)} = 10^2 = 100\text{ V}$$

RESULT: The potential difference between points A and B is $100\text{ volts}$.

Ohm's Law & Circuit Resistance

5. The potential difference applied to a portable radio terminal is $6.0\text{ Volts}$. Determine the resistance of the radio when a current of $20\text{ mA}$ flows through it.

Data:

• Potential difference ($V$) = $6.0\text{ V}$
• Current ($I$) = $20\text{ mA} = 20 \times 10^{-3}\text{ A} = 0.02\text{ A}$
• Resistance ($R$) = ?

Solution:

According to Ohm's Law:

$$V = I \cdot R \implies R = \frac{V}{I}$$ $$R = \frac{6.0\text{ V}}{0.02\text{ A}}$$ $$R = 300\text{ }\Omega$$

RESULT: The total dynamic resistance of the radio is $300\text{ }\Omega$.

6. Resistances of $4\text{ }\Omega$, $6\text{ }\Omega$, and $12\text{ }\Omega$ are connected in parallel and then connected to a $6\text{V}$ EMF source. Determine the value of:
i) The circuit's equivalent resistance.
ii) The total current flowing through the circuit.
iii) The branch current that flows through each individual resistance.

Data:

• Resistance ($R_1$) = $4\text{ }\Omega$
• Resistance ($R_2$) = $6\text{ }\Omega$
• Resistance ($R_3$) = $12\text{ }\Omega$
• Source Voltage ($V$) = $6\text{ V}$

Solution:

i) Finding the Equivalent Resistance ($R_{eq}$):

For a parallel resistor grouping network:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ $$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12}$$ $$\text{Taking LCM of 4, 6, and 12, which is 12:}$$ $$\frac{1}{R_{eq}} = \frac{3 + 2 + 1}{12} = \frac{6}{12} = \frac{1}{2}$$ $$R_{eq} = 2\text{ }\Omega$$

ii) Finding Total Current ($I_{\text{total}}$):

Using the system Ohm's Law formula:

$$I_{\text{total}} = \frac{V}{R_{eq}} = \frac{6\text{ V}}{2\text{ }\Omega} = 3\text{ A}$$

iii) Finding Branch Currents ($I_1, I_2, I_3$):

Since the voltage across all components is equal in parallel configurations ($V = 6\text{ V}$):

$$I_1 = \frac{V}{R_1} = \frac{6\text{ V}}{4\text{ }\Omega} = 1.5\text{ A}$$ $$I_2 = \frac{V}{R_2} = \frac{6\text{ V}}{6\text{ }\Omega} = 1.0\text{ A}$$ $$I_3 = \frac{V}{R_3} = \frac{6\text{ V}}{12\text{ }\Omega} = 0.5\text{ A}$$

RESULT:
i) The circuit's equivalent resistance is $2\text{ }\Omega$.
ii) The total system current is $3\text{ A}$.
iii) The branch currents flowing through the $4\text{ }\Omega$, $6\text{ }\Omega$, and $12\text{ }\Omega$ resistors are $1.5\text{ A}$, $1.0\text{ A}$, and $0.5\text{ A}$ respectively.