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QuestionThe circuit has been connected as shown in he figure for a "long" time.
What is the magnitude of the electric potential EC\mathcal{E}_{C} across the capacitor?

Studdy Solution

STEP 1

What is this asking? After the circuit's been on for a while, what's the voltage across the capacitor? Watch out! Capacitors act like open circuits after a long time, so no current flows through them!

STEP 2

1. Simplify the Resistor Network
2. Calculate the Current
3. Find the Voltage

STEP 3

Alright class, let's **rock and roll** with this circuit problem!
Since the capacitor acts like an open circuit after a long time, we can ignore it for now.
The 16 Ω\text{16 } \Omega and 4 Ω\text{4 } \Omega resistors are in series, so their combined resistance is simply their sum.

STEP 4

R1=16 Ω+4 Ω=20 Ω R_{1} = 16\ \Omega + 4\ \Omega = \textbf{20}\ \Omega Awesome! Now, this **new** 20 Ω\textbf{20}\ \Omega resistance is in parallel with the *existing* 20 Ω\text{20 } \Omega resistor.

STEP 5

R2=1120 Ω+120 Ω=1220 Ω=20 Ω2=10 Ω R_{2} = \frac{1}{\frac{1}{20\ \Omega} + \frac{1}{20\ \Omega}} = \frac{1}{\frac{2}{20\ \Omega}} = \frac{20\ \Omega}{2} = \textbf{10}\ \Omega Boom! We've simplified the parallel resistors down to a single 10 Ω\textbf{10 } \Omega resistance.

STEP 6

This 10 Ω\textbf{10 } \Omega resistance is now in series with the 8 Ω\text{8 } \Omega resistor, giving us a **total** equivalent resistance: Req=10 Ω+8 Ω=18 Ω R_{eq} = 10\ \Omega + 8\ \Omega = \textbf{18}\ \Omega

STEP 7

Now that we have the **total equivalent resistance**, we can find the **total current** flowing through the circuit using Ohm's Law, V=I⋅RV = I \cdot R.

STEP 8

I=VReq=6 V18 Ω=13 A I = \frac{V}{R_{eq}} = \frac{6\ V}{18\ \Omega} = \frac{\textbf{1}}{\textbf{3}}\ A So, we have 13\frac{\textbf{1}}{\textbf{3}} amps flowing from the battery.

STEP 9

Remember that 10 Ω\textbf{10 } \Omega resistance we calculated earlier?
That's the combined resistance of the two 20 Ω\text{20 } \Omega resistors *and* the 16 Ω\text{16 } \Omega and 4 Ω\text{4 } \Omega resistors.
The voltage across this 10 Ω\textbf{10 } \Omega equivalent resistance is what we're after, because *that's* the voltage across the capacitor!

STEP 10

Using Ohm's Law again: VC=I⋅R2=13 A⋅10 Ω=103 V V_{C} = I \cdot R_{2} = \frac{1}{3}\ A \cdot 10\ \Omega = \frac{\textbf{10}}{\textbf{3}}\ V

STEP 11

The voltage across the capacitor is 103\frac{10}{3} V.

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