Electric Field
Resistance in electric circuit

Resistance in an electric circuit is the opposition that a material or component offers to the flow of electric current. It is measured in ohms (Ω) and is influenced by factors such as the material's type, length, cross-sectional area, and temperature. The higher the resistance, the lower the current flow for a given voltage, according to Ohm’s Law: V = IR. It is measured using a resistor

Imagine electricity is like water flowing through a pipe. Resistance is like something in the pipe that slows the water down, like a sponge or a narrow spot. In a wire, resistance makes it harder for electricity to move. The more resistance there is, the slower the electricity flows.

resistors connection
  1. Series Connection:

    In a series connection, resistors are connected end to end. The same current flows through each resistor, but the total voltage is shared across them.

    • Total Resistance: \( R_{\text{total}} = R_1 + R_2 + R_3 + \cdots \)
    • Total Current: \( I_T = I_1 = I_2 = \cdots \) (same through all resistors)
    • Voltage across each resistor: \( V_n = I \times R_n \)
    • Total Voltage: \( V_T = V_1 + V_2 + V_3 + \cdots \)


  2. Parallel Connection:

    In a parallel connection, resistors are connected side by side. The same voltage is applied across each resistor, but the total current is divided among them.

    • Total Resistance: \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots \)
    • Voltage: \( V \) (same across all resistors)
    • Current through each resistor: \( I_n = \frac{V}{R_n} \)
    • Total Current: \( I_{\text{total}} = I_1 + I_2 + I_3 + \cdots \)

Ohms law

Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference (voltage) across it, provided the temperature and other physical conditions remain constant. This means if you increase the voltage, the current increases at the same rate.

Mathematically, it is expressed as \( V = IR \),
where \( V \) is voltage in volts,
\( I \) is current in amperes, and
\( R \) is resistance in ohms (Ω).

The graph of Ohm’s Law is a straight line when voltage \( V \) is plotted on the y-axis and current \( I \) on the x-axis. The slope of the line represents the resistance \( R \), showing a linear relationship between voltage and current.

Calculations

Example 1: A parallel combination of \( 3 \ \Omega \) and 4 \( \Omega \) resistors is connected in series with a resistor of \( 4 \Omega \) and a battery of negligible resistance. Calculate the effective resistance in the circuit. (WAEC)

Solution

$$ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} $$ $$ \frac{1}{R_{\text{parallel}}} = \frac{1}{3} + \frac{1}{4} $$ $$ \frac{1}{R_{\text{parallel}}} = \frac{4+3}{12} $$ $$ R_{\text{parallel}} = \frac{12}{7} ≈ 1.71 \Omega $$ $${\text{Now both resistors are in series}} $$ $$ R_{\text{total}} = 1.71 + 4 $$ $$ R_{\text{total}} = 5.71 \Omega $$

Example 2: Two resistors of 3Ω and 6Ω are connected in parallel. Find the total resistance. (NECO)

Solution

$$ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} $$ $$ \frac{1}{R_{\text{total}}} = \frac{1}{3} + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} $$ $$ R_{\text{total}} = \frac{6}{3} = 2\ \Omega $$

Example 3: A radio is operated by eight cells each of e.m.f 2.0V connected in series. If two of the cells are wrongly connected, find the net e.m.f of the radio. (WAEC)

Solution

Total number of cells = 8

E.m.f of each cell = 2.0V

Two cells are wrongly connected, so their e.m.f will subtract from the rest.

$$ \text{Correctly connected cells} = 8 - 2 = 6 $$ $$ \text{Net e.m.f} = (8 \times 2.0) - (2 \times 2.0) $$ $$ = 16.0V - 4.0V $$ $$ = 12.0V $$

Example 4: Four resistors of 5Ω each are connected in series. Calculate the voltage needed to produce a current of 2A. (JAMB)

Solution

$$ R_{\text{total}} = 5 + 5 + 5 + 5 = 20\ \Omega $$ $$ I = 2A $$ $$ V = IR = 2 × 20 = 40V $$

Example 5: Two resistors of 8Ω and 12Ω are connected in parallel. What current flows through the combination if the voltage across them is 24V? (WAEC)

Solution

$$ \frac{1}{R_{\text{total}}} = \frac{1}{8} + \frac{1}{12} = \frac{3 + 2}{24} = \frac{5}{24} $$ $$ R_{\text{total}} = \frac{24}{5} = 4.8\ \Omega $$ $$ V = 24V $$ $$ I = \frac{V}{R_{\text{total}}} = \frac{24}{4.8} = 5A $$

Example 6: A 24V potential difference is applied across a parallel combination of four 6Ω resistors. Find the current in each resistor. (JAMB)

Solution

In a parallel circuit, the voltage across each resistor is the same as the total voltage.

$$ V = 24V $$

Each resistor has a resistance of:

$$ R = 6\ \Omega $$

Using Ohm’s Law:

$$ I = \frac{V}{R} = \frac{24}{6} = 4A $$

Toal current in the curcuit = 16A

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Summary