Part A = counterclockwise

Part B = I = 2EMF / ((2 * R_{1}) + R_{2})

Part C = I_{B} = 2*EMF / (R_{1} + 2*R_{2})

Part D = 0.064 W

Part E = R_{1} / R_{2} = 1

Part F = R_{2} > R_{1}

In which direction does the current in circuit A flow?

What is the current through the resistor of resistance R_{2} in circuit A?

Calculate the current I_{B} through the resistor of resistance R_{2} for circuit B.

What is the power dissipated by the resistor of resistance R_{2} for circuit A, given that Ε = 10 V, R_{1} = 300 ohms, and R_{2} = 5000 ohms?

For what ratio of R_{1} and R_{2} would power dissipated by the resistor of resistance R_{2} be the same for circuit A and circuit B?

Under which of the following conditions would power dissipated by the resistance R_{2} in circuit A be bigger than that of circuit B?

Part A = current

Part B = I_{2} + I_{3} – I_{1}

Part C = I_{3} ⋅ R_{3} – I_{2} ⋅ R_{2}

Part D = V_{b} – I_{1} ⋅ R_{1} – I_{3} ⋅ R_{3}

The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a steady state.

Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R_{2}).

Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal.

Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow.

Part A = Zero

Part B = Ε

Part C = clockwise

Part D = the top plate

Part E = zero

Part F = q = CΕ

Part G = W = CΕ^2

Part H = q(t) = CΕ * (1 – e^(-t/RC))

Part I = I(t) = (Ε / R) * e^(-t/RC)

Part J = I(t) = (Ε / R) * e^(-t/RC)

Part K = I(t) = -q_{0} * e^(-t/RC) / (RC)

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Part A = 9.0 v

Part B = 1.6 v

Part C = 0.39 v

Part D = Increase

What does the voltmeter read on a sunny day?

What does the voltmeter read on a cloudy day?

What does the voltmeter read at night?

Does the voltmeter reading increase or decrease as the light intensity increases?

Part A = P_{R}(t) = I(t)^{2}R

Part B = P_{L}(t) = -I(t)^{2}R

Part C = Power comes out of the inductor and is dissipated by the resistor

Consider an R-L circuit with a DC voltage source, as shown in the figure. This circuit has a current I_{c} when t < 0. At t = 0 the switch is thrown removing the DC voltage source from the circuit. The current decays to I(t) at time t.

What is the power, P_{R}(t), flowing into the resistor, R, at time t?

What is the power flowing into the inductor?

Compare the two equations for power dissipated within the resistor and inductor. Which of the following conclusions about the shift of energy within the circuit can be made?

If the external resistance is then changed to R_{2} = 4.00 Ω, what is the value of the current I_{2} in the circuit?