Answer :
The potential difference of 3 V occurs in the circuit with the 12-ohm resistor.
To determine in which circuit the bulb experiences a potential difference of 3 V, we need to analyze both circuits step-by-step.
Circuit 1:
It has a 9 V battery and a 12-ohm resistor connected in series with the 6-ohm bulb.
The total resistance in the circuit can be calculated as:
[tex]R_{total} = R_{bulb} + R_{resistor} = 6 \Omega + 12 \Omega = 18 \Omega[/tex]
According to Ohm's Law, the current (I) flowing through the circuit can be found using:
[tex]I = \frac{V}{R} = \frac{9 V}{18 \Omega} = 0.5 A[/tex]
To find the potential difference (V) across the bulb, we can again use Ohm's Law, now for just the bulb:
[tex]V_{bulb} = I \times R_{bulb} = 0.5 A \times 6 \Omega = 3 V[/tex]
Circuit 2:
This circuit consists only of the 9 V battery and the 6-ohm bulb without the 12-ohm resistor.
The total resistance in this circuit is simply the resistance of the bulb:
[tex]R_{total} = 6 \Omega[/tex]
Again using Ohm's Law for the current, we find:
[tex]I = \frac{9 V}{6 \Omega} = 1.5 A[/tex]
Now, to find the potential difference across the bulb:
[tex]V_{bulb} = I \times R_{bulb} = 1.5 A \times 6 \Omega = 9 V[/tex]
Conclusion:
From our calculations, we see that in Circuit 1, the bulb receives a potential difference of 3 V, while in Circuit 2, it receives a 9 V potential difference.
Thus, the correct answer is that the potential difference of 3 V occurs in the circuit with the 12-ohm resistor.
Answer:
9 V
Explanation:
For the first circuit:
[tex]R_{total}=R+R_b=12+6=\bf18\:$\Omega$\\[/tex]
We apply Ohm's Law:
[tex]I=\frac{V}{R_{total}}=\frac{9}{18}=\frac{1}{2}=0.5\:A[/tex]
We find the voltage for the bulb:
[tex]V_b=I+R_b=0.5\times6=\bf3\:V\\[/tex]
For the second circuit:
We apply Ohm's Law:
[tex]I=\frac{V}{R_b}=\frac{9}{6}=\frac{3}{2}=1.5\:A\\[/tex]
We find the voltage for the bulb:
[tex]V_b=I\times R_b=1.5\times6=\bf9\:V[/tex]
