If the first (horizontal) resistor had zero resistance, what would be the initial voltage across the capacitor? Express your answer in terms of E.
At least I know why the answer I typed into the problem, which was V_C,0 = E, was definitely incorrect, because my answer disagrees with Kirchhoff's loop rule.
However, I need help on writing the rule for a loop that contains a battery and a capacitor, and I'm stuck on it.
Well, it seems like you're trying to figure out the voltage across a capacitor when the first resistor has zero resistance. That's quite a shock! In this case, we can say goodbye to the first resistor and focus on the elements that remain.
To write the loop rule for a loop that contains a battery and a capacitor, we need to consider the voltage drops and rises across the components.
Since the first resistor has zero resistance, it won't contribute any voltage drop. So, we can skip it and focus on the battery and the capacitor instead.
The loop rule tells us that the sum of the voltage drops must be equal to the sum of the voltage rises in a closed loop. In this case, the only voltage drop is across the capacitor and the voltage rise is from the battery.
So, we can write the loop rule as:
-ΔVc = ΔVbattery
But what's ΔVc? Well, ΔVc represents the change in voltage across the capacitor. Since we're interested in the initial voltage, we can express ΔVc as:
ΔVc = Vc,0 - Vc
Now, we know that initially, when the first resistor has zero resistance, the capacitor is uncharged. So, Vc,0 = 0.
Substituting these values into our loop rule equation, we get:
0 - Vc = ΔVbattery
But, we know that the voltage across the battery, which is ΔVbattery, is simply E (the electromotive force).
Therefore, the initial voltage across the capacitor would be:
Vc = -E
Keep in mind that the negative sign just represents the orientation of the voltage drop across the capacitor in relation to the loop direction. So, don't go negative on yourself for finding a negative voltage!
Hope that puts a charge in your understanding!
When writing the Kirchhoff's loop rule for a loop that contains a battery and a capacitor, we need to consider the potential difference across the battery and the potential difference across the capacitor.
Let's denote the potential difference across the battery as V_B and the potential difference across the capacitor as V_C.
The Kirchhoff's loop rule states that the sum of the potential differences around any closed loop is equal to zero.
In this specific case, considering a clockwise loop starting from the positive terminal of the battery, we can write the loop equation as:
V_B - V_C = 0
Since the first resistor has zero resistance, no voltage drop occurs across it. Therefore, the initial voltage across the capacitor, V_C,0, is equal to the potential difference across the battery, which is E.
So, the correct initial voltage across the capacitor, when the first resistor has zero resistance, is V_C,0 = E.
When analyzing circuits with batteries and capacitors, it is important to consider Kirchhoff's loop rule, which states that the sum of the voltages around any closed loop in a circuit must be zero.
In a loop that contains a battery and a capacitor, we need to consider the equation relating the voltage across a capacitor to the charge stored on it. The equation is:
Q = C * V
Where:
Q is the charge stored on the capacitor,
C is the capacitance of the capacitor, and
V is the voltage across the capacitor.
To write the loop rule, we need to consider the direction of the loop. Let's start at a point, move around the loop, and return to the starting point.
1. Moving from the starting point towards the battery:
- We move from a point of lower potential to higher potential due to the battery's positive terminal.
- The voltage across the battery is E, and the voltage across the capacitor is V.
- Thus, the voltage change in this segment of the loop is E - V.
2. Moving through the battery:
- The voltage change across the battery is E, and it is in the opposite direction compared to the previous segment.
- Therefore, the voltage change in this segment of the loop is -E.
3. Moving from the battery towards the capacitor:
- We move from a point of higher potential to lower potential due to the capacitor.
- The voltage change in this segment is -V because we move from a higher voltage (across the capacitor) to a lower voltage.
Now, applying Kirchhoff's loop rule:
(E - V) - E - V = 0
Simplifying the equation:
-E + E - V - V = 0
-2V = 0
Since we have assumed that the first resistor has zero resistance, it implies that there is no voltage drop across it. Therefore, the voltage across the capacitor, V, should be zero.
Hence, the initial voltage across the capacitor, when the first resistor has zero resistance, is 0.