Electricity - Calculate resistance, resistivity, and equivalent resistance in series and parallel circuits

A wire of resistance $10 \Omega$ is stretched so that its length becomes three times its original length. If the volume remains constant, calculate the new resistance.

Let initial length be $l$ and area be $A$. Resistance $R = \rho \frac{l}{A} = 10 \Omega$. When stretched to $3l$, the area becomes $\frac{A}{3}$ (since Volume $V = l \times A$ is constant). New resistance $R' = \rho \frac{3l}{A/3} = 9 \left( \rho \frac{l}{A} \right) = 9 \times 10 = 90 \Omega$.

Stretching a wire increases its length and simultaneously decreases its cross-sectional area. Because $R$ is proportional to $l$ and inversely proportional to $A$, the resistance increases by the square of the change in length.

Compare the resistance of two wires of the same material: Wire A has length $L$ and radius $r$, Wire B has length $2L$ and radius $2r$.

Resistance of Wire A: $R_A = \rho \frac{L}{\pi r^2}$. Resistance of Wire B: $R_B = \rho \frac{2L}{\pi (2r)^2} = \rho \frac{2L}{4 \pi r^2} = \frac{1}{2} \left( \rho \frac{L}{\pi r^2} \right) = \frac{1}{2} R_A$.

Even though Wire B is twice as long (which increases resistance), its radius is doubled, making its area four times larger (which decreases resistance). The net effect is that Wire B has half the resistance of Wire A.

Calculate the equivalent resistance of the circuit shown, where $R_1 = 4 \Omega$, $R_2 = 8 \Omega$, and $R_3 = 8 \Omega$.

First, calculate the parallel combination of $R_2$ and $R_3$: $\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3}$ $\frac{1}{R_p} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$ So, $R_p = 4 \Omega$. Now, the total resistance $R_{eq}$ is the series sum of $R_1$ and $R_p$: $R_{eq} = R_1 + R_p$ $R_{eq} = 4 + 4 = 8 \Omega$

We simplify the circuit by first solving the parallel branch ($R_2$ and $R_3$) to find its single equivalent resistance, then adding that in series to $R_1$.

Determine the current flowing through a $2 \Omega$ resistor when it is connected in parallel with a $3 \Omega$ and $6 \Omega$ resistor, all connected to a $12 V$ battery.

In a parallel circuit, the voltage across each resistor is equal to the supply voltage. Thus, $V = 12 V$ for each resistor. Using Ohm's law for the $2 \Omega$ resistor ($R_1$): $I_1 = \frac{V}{R_1}$ $I_1 = \frac{12}{2} = 6 A$

Because the resistors are in parallel, each resistor experiences the full potential difference of the battery. The current through a specific resistor is independent of the others.