Question

0 ω resistor is in series with a bulb and a 12 v source. if 8 v is across the bulb, what voltage is across the resistor? what is the current in the circuit? what is the resistance of the bulb?

Explanation:

Step1: Find voltage across resistor

In a series - circuit, the sum of voltages across components equals the source voltage. Let $V_s = 12V$ be the source voltage, $V_b=8V$ be the voltage across the bulb, and $V_r$ be the voltage across the resistor. Then $V_s=V_b + V_r$. So, $V_r=V_s - V_b$.
$V_r=12 - 8=4V$

Step2: Find current in the circuit

We know the voltage across the resistor $V_r = 4V$ and assume the resistance of the resistor $R_r = 10\Omega$ (the value of the resistor is not given in the problem statement, but we need a value to calculate the current. Let's assume $R_r = 10\Omega$ for the sake of calculation. If the actual value is different, the following steps can be adjusted accordingly). Using Ohm's law $I=\frac{V}{R}$, for the resistor, $I = \frac{V_r}{R_r}$.
$I=\frac{4}{10}=0.4A$

Step3: Find resistance of the bulb

Using Ohm's law for the bulb, $R_b=\frac{V_b}{I}$. Since $V_b = 8V$ and $I = 0.4A$, then $R_b=\frac{8}{0.4}=20\Omega$

Answer:

Voltage across the resistor: $4V$
Current in the circuit: $0.4A$
Resistance of the bulb: $20\Omega$