Electricity - Define electric current, potential difference, and state Ohm's law

A current of $0.5 \text{ A}$ is drawn by a filament of an electric bulb for $10$ minutes. Find the amount of electric charge that flows through the circuit.

Given: $I = 0.5 \text{ A}$, $t = 10 \text{ minutes} = 10 \times 60 = 600 \text{ s}$. Using $Q = I \times t$: $Q = 0.5 \text{ A} \times 600 \text{ s} = 300 \text{ C}$.

To find the charge, we multiply the current by the time in seconds. $10$ minutes must be converted to $600$ seconds for standard SI units.

How much work is done in moving a charge of $2 \text{ C}$ across two points having a potential difference of $12 \text{ V}$?

Given: $Q = 2 \text{ C}$, $V = 12 \text{ V}$. Using $W = V \times Q$: $W = 12 \text{ V} \times 2 \text{ C} = 24 \text{ J}$.

Work done is the product of the potential difference and the amount of charge moved.

Calculate the number of electrons constituting one Coulomb of charge.

Given: $Q = 1 \text{ C}$, $e = 1.6 \times 10^{-19} \text{ C}$. Using $n = \frac{Q}{e}$: $n = \frac{1}{1.6 \times 10^{-19}} \approx 6.25 \times 10^{18}$ electrons.

By applying the quantization of charge formula $Q = ne$, we find the total number of elementary charges required to sum up to one Coulomb.

The potential difference between the terminals of an electric heater is $60 \text{ V}$ when it draws a current of $4 \text{ A}$ from the source. What current will the heater draw if the potential difference is increased to $120 \text{ V}$?

Step 1: Find Resistance ($R$) using the first case. Given $V_1 = 60 \text{ V}$, $I_1 = 4 \text{ A}$. According to Ohm's law, $R = \frac{V_1}{I_1}$ $R = \frac{60 \text{ V}}{4 \text{ A}} = 15 \text{ } \Omega$

Step 2: Find new current ($I_2$) when $V_2 = 120 \text{ V}$. $I_2 = \frac{V_2}{R}$ $I_2 = \frac{120 \text{ V}}{15 \text{ } \Omega} = 8 \text{ A}$ The current through the heater becomes $8 \text{ A}$.

Resistance remains constant for the same heater. Since $V$ is doubled ($60 \text{ V}$ to $120 \text{ V}$), the current $I$ also doubles ($4 \text{ A}$ to $8 \text{ A}$) following $V \propto I$.

Resistance of an electric iron is $50 \text{ } \Omega$. If a current of $4.4 \text{ A}$ flows through it, calculate the potential difference between the two terminals of the electric iron.

Given: Resistance $R = 50 \text{ } \Omega$ Current $I = 4.4 \text{ A}$

Using Ohm's Law: $V = I \times R$ $V = 4.4 \text{ A} \times 50 \text{ } \Omega$ $V = 220 \text{ V}$

The potential difference is $220 \text{ V}$.

By multiplying the flow of charge (current) by the opposition to that flow (resistance), we find the electrical pressure (potential difference) required.