A galvanometer is a device used to detect small electric currents, typically in the range of a few hundred microamperes (μA). It is very sensitive and can show even very small deflections when current flows through it.

To measure larger currents without damaging the galvanometer, a shunt resistor is added in parallel with the galvanometer. This configuration forms an ammeter.

1. The shunt resistor allows most of the current to bypass the galvanometer, thus preventing excessive current from flowing through it.
2. Only a small fraction of the current flows through the galvanometer, protecting it from potential damage due to high current.

Formula for voltage across an Ohmic device

In an Ohmic device, where the resistance is constant, the voltage $V$ across the device is related to the current $I$ by the formula:

$V = I \times R$

Where:

• $V$ is the potential difference (voltage) across the device (in volts),
• $I$ is the current flowing through the device (in amperes),
• $R$ is the resistance of the device (in ohms, $\Omega$).

Combination of resistors

There are two main ways to connect resistors:

1. In Series
2. In Parallel

Series connection

In a series connection, resistors are connected end-to-end. The same current flows through all resistors, but the voltage drop varies across each resistor.

If three resistors $R_1$, $R_2$, and $R_3$ are connected in series, the voltage across each resistor is given by:

$V_1 = I \times R_1$

$V_2 = I \times R_2$

$V_3 = I \times R_3$

The total voltage in the series connection is the sum of the voltage drops across each resistor:

$V = V_1 + V_2 + V_3 = I \times (R_1 + R_2 + R_3)$

The equivalent resistance $R_{eq}$ of resistors connected in series is the sum of their individual resistances:

$R_{eq} = R_1 + R_2 + R_3$

Example:

Given resistors: $5 \, \Omega$, $10 \, \Omega$, and $5 \, \Omega$

Battery voltage: $15 \, \text{V}$

Total resistance:

$R = 5 + 10 + 5 = 20 \, \Omega$

Current:

$I = \frac{V}{R} = \frac{15}{20} = 0.75 \, \text{A}$

Voltage drop across each resistor:

$V = I \times R = 0.75 \times (5 + 10 + 5) = 15 \, \text{V}$

Parallel connection

In a parallel connection, resistors are connected across the same two points. The voltage is the same across each resistor, but the current varies depending on the resistance.

The total current in the circuit is the sum of the currents through each resistor. For resistors $R_1$, $R_2$, and $R_3$ connected in parallel, the total current is given by:

$I = I_1 + I_2 + I_3$

The reciprocal of the total (equivalent) resistance $R_{eq}$ for resistors connected in parallel is given by:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Example:

Given resistors: $6 \, \Omega$, $12 \, \Omega$, and $18 \, \Omega$

Total equivalent resistance:

$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{18}$

First, calculate the sum of the reciprocals:

$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{18} = 0.4167$

Now, take the reciprocal to find $R_{eq}$:

$R_{eq} = \frac{1}{0.4167} = 2.4 \, \Omega$

Resistor connections and calculations

If three resistors $R_1$, $R_2$, $R_3$ are connected in parallel:

Total current:

$I = I_1 + I_2 + I_3$

Current through each resistor:

$I_1 = \frac{V}{R_1}, \, I_2 = \frac{V}{R_2}, \, I_3 = \frac{V}{R_3}$

Equivalent resistance:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Note: The equivalent resistance in parallel is always less than the smallest resistor.

Example

Given resistors: $5\Omega$, $10\Omega$, $5\Omega$ and battery voltage $15V$:

Total resistance:

$\frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{5} = \frac{5}{10} \Rightarrow R = 2\Omega$

Current through each:

$I_1 = \frac{15}{5} = 3A, \, I_2 = \frac{15}{10} = 1.5A, \, I_3 = \frac{15}{5} = 3A$

Total current:

$I = 3 + 1.5 + 3 = 7.5A$

When some resistors are in parallel and others in series, we:

First, calculate the equivalent resistance for the parallel part:

$\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} \Rightarrow R_{23} = \frac{R_2 R_3}{R_2 + R_3}$

Then add the series resistor:

$R_{eq} = R_1 + R_{23}$

Total current:

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

Formula for total current in combined circuit:

$I = \frac{V}{(R_2 + R_3)} \left( \frac{R_1 R_2}{R_1 R_2 + R_1 R_3 + R_2 R_3} \right)$

• Each device gets full voltage from the source.
• Switching one device on/off doesn't affect others.

Example 3

Given:

Battery voltage = 24V

Two resistors: $4\Omega$ and $6\Omega$ in series

Total resistance:

$R = 4\Omega + 6\Omega = 10\Omega$

Current:

$I = \frac{24V}{10\Omega} = 2.4A$

p.d across $4\Omega$:

$V = I \times R = 2.4 \times 4 = 9.6V$

p.d across $6\Omega$:

$24V - 9.6V = 14.4V$

Therefore:

p.d across $4\Omega$ = 9.6V

p.d across $6\Omega$ = 14.4V