Question

what is the resistance of a light bulb if a potential difference of 120 v will produce a current of 0.5 a in the bulb?
○ 0.0042 ω
○ 0.5 ω
○ 60 ω
○ 240 ω

Explanation:

Step1: Recall Ohm's Law

Ohm's Law is given by \( V = IR \), where \( V \) is the potential difference (voltage), \( I \) is the current, and \( R \) is the resistance. We need to solve for \( R \), so we rearrange the formula to \( R=\frac{V}{I} \).

Step2: Substitute the given values

We are given \( V = 120\ V \) and \( I = 0.5\ A \). Substituting these values into the formula \( R=\frac{V}{I} \), we get \( R=\frac{120}{0.5} \).

Step3: Calculate the resistance

Calculating \( \frac{120}{0.5} \), we know that dividing by 0.5 is the same as multiplying by 2, so \( 120\times2 = 240 \). So the resistance \( R = 240\ \Omega \).

Answer:

240 Ω (corresponding to the option "240 Ω")