Electricity - CIRCUIT DIAGRAM11.3 CIRCUIT DIAGRAM11.3 CIRCUIT DIAGRAM11.3 CIRCUIT DIAGRAM11.3 CIRCUIT DIAGRAM

A circuit contains a battery of $3$ cells of $2\text{ V}$ each, a $5\text{ }\Omega$ resistor, an $8\text{ }\Omega$ resistor, and a $12\text{ }\Omega$ resistor, all connected in series with a closed plug key and an ammeter. Calculate the reading shown by the ammeter.

Total Potential Difference $(V) = 3 \times 2\text{ V} = 6\text{ V}$. Since resistors are in series, Total Resistance $(R_s) = 5\text{ }\Omega + 8\text{ }\Omega + 12\text{ }\Omega = 25\text{ }\Omega$. Using Ohm's Law: $I = \frac{V}{R}$ $I = \frac{6}{25} = 0.24\text{ A}$

In a series circuit, the total resistance is the sum of individual resistances. The ammeter measures the total current flowing through the circuit, which is calculated using the total voltage and total resistance.

Calculate the equivalent resistance of two resistors of $10\text{ }\Omega$ and $40\text{ }\Omega$ when they are connected in parallel in a circuit diagram.

Given $R_1 = 10\text{ }\Omega$ and $R_2 = 40\text{ }\Omega$. Formula for parallel resistance: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$ $\frac{1}{R_p} = \frac{1}{10} + \frac{1}{40}$ $\frac{1}{R_p} = \frac{4 + 1}{40} = \frac{5}{40}$ $R_p = \frac{40}{5} = 8\text{ }\Omega$

When resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.

Draw a circuit diagram representing a battery of four cells of $1.5\text{ V}$ each, a $10\text{ }\Omega$ resistor, a $20\text{ }\Omega$ resistor, and an ammeter, all connected in series with a closed switch. Calculate the total potential difference and the reading shown by the ammeter.

1. Total potential difference ($V$): Since there are four cells of $1.5\text{ V}$ each in series, $V = 4 \times 1.5\text{ V} = 6\text{ V}$.
2. Total resistance ($R_s$): Since the resistors are in series, $R_s = R_1 + R_2 = 10\text{ }\Omega + 20\text{ }\Omega = 30\text{ }\Omega$.
3. Current ($I$): Using Ohm's Law, $I = \frac{V}{R_s} = \frac{6\text{ V}}{30\text{ }\Omega} = 0.2\text{ A}$. The ammeter reading is $0.2\text{ A}$.

In a series circuit, the total voltage is the sum of the individual cell voltages, and the total resistance is the sum of individual resistances. The ammeter measures the same current flowing through all components in a series loop.

Consider a circuit where two resistors of $6\text{ }\Omega$ and $3\text{ }\Omega$ are connected in parallel to a $12\text{ V}$ battery. A voltmeter is connected across the $6\text{ }\Omega$ resistor. Determine the reading of the voltmeter and the total current drawn from the battery.

1. Voltmeter reading: In a parallel circuit, the potential difference across each branch is the same as the supply voltage. Therefore, the voltmeter reading across the $6\text{ }\Omega$ resistor is $12\text{ V}$.
2. Equivalent resistance ($R_p$): $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$, so $R_p = 2\text{ }\Omega$.
3. Total current ($I$): $I = \frac{V}{R_p} = \frac{12\text{ V}}{2\text{ }\Omega} = 6\text{ A}$.

A voltmeter is always connected in parallel. In a parallel arrangement, the voltage remains constant across all branches, while the total current is the sum of the currents through each branch (or calculated via equivalent resistance).