A)Two identical resistors are connected in parallel across a 30-V battery, which supplies them with a total power of 7.0 W. While the battery is still connected, one of the resistors is heated so that its resistance doubles. The resistance of the other resistor remains unchanged. Find (a) the initial resistance of each resistor, and (b) the total power delivered to the resistors after one resistor has been heated.

B)You have three capacitors: C1 = 60 ìF, C2 = 48 ìF, and C3 = 36 ìF. Determine the maximum equivalent capacitance you can obtain by connecting two of the capacitors in parallel and then connecting the parallel combination in series with the remaining capacitor.

C)A 2.64-F and a 7.03-F capacitor are connected in series across a 30.0-V battery. A 7.44-F capacitor is then connected in parallel across the 2.64-F capacitor. Determine the voltage across the 7.44-F capacitor.

100

A.

a. E^2/R = 7
30^2/R = 7
R = 128.6 ohms.
R1 = R2 = 2*128.6 = 257.2 ohms.

b. 1/R = 1/257 + 1/2*257 = 0.005837
R = 171 ohms.
P = E^2/R = 30^2/171 = 5.2 Watts.

A) To solve for the initial resistance of each resistor, we can use the formula for power in a circuit:

P = V^2 / R,

where P is the power, V is the voltage, and R is the resistance. Since the power is given as 7.0 W and the voltage is given as 30 V, we can write:

7.0 = (30^2) / (R1 + R2),

where R1 and R2 are the resistances of the two identical resistors.

Next, we need to find the relationship between the initial and final resistances. We know that one resistor's resistance doubles while the other remains unchanged. Let's call the initial resistance of one resistor R and the initial resistance of the other resistor 2R.

So, after heating one resistor, the new resistance values are R and 4R.

Now, we can substitute these relationships into the equation and solve for the initial resistance:

7.0 = (30^2) / (R + 2R),
7.0 = (30^2) / (3R),
R = (30^2) / (7.0 * 3).

The initial resistance of each resistor is R ≈ 85.71 Ω.

(b) To find the total power delivered to the resistors after one resistor has been heated, we need to calculate the new resistance of the system. The heated resistor has doubled in resistance, so now we have R and 2R as the resistances.

To find the total resistance in a parallel circuit, we can use the formula:

1/Rtotal = 1/R1 + 1/R2.

Substituting the values, we get:

1/Rtotal = 1/R + 1/(2R),
1/Rtotal = 3/(2R),
Rtotal = 2R/3.

Now we can calculate the total power using the equation P = V^2 / Rtotal:

Ptotal = (30^2) / (2R/3).

The total power delivered to the resistors after one resistor has been heated is Ptotal ≈ 315 W.

B) To determine the maximum equivalent capacitance by connecting two capacitors in parallel and then connecting the parallel combination in series with another capacitor, we need to use the formulas for capacitors in parallel and series.

When capacitors are connected in parallel, their capacitances add up:

Cparallel = C1 + C2.

In this case, connecting C1 and C2 in parallel gives us a new capacitor Cparallel = 60 μF + 48 μF = 108 μF.

Now, we need to connect this parallel combination in series with C3. When capacitors are connected in series, their reciprocals add up:

1/Cseries = 1/Cparallel + 1/C3.

Substituting the values, we get:

1/Cseries = 1/108 μF + 1/36 μF,
1/Cseries = (3 + 9) / (108 μF),
1/Cseries = 12 / (108 μF),
Cseries = (108 μF) / 12.

The maximum equivalent capacitance you can obtain by connecting two capacitors in parallel and then connecting the combination in series with the remaining capacitor is Cseries ≈ 9 μF.

C) To determine the voltage across the 7.44-F capacitor, we need to analyze the circuit step by step.

First, we have the initial series circuit with the 2.64-F and 7.03-F capacitors connected across the 30.0-V battery. The voltage across each capacitor in a series circuit is the same.

So, the voltage across the 2.64-F and 7.03-F capacitors is 30.0 V.

Next, we connect the 7.44-F capacitor in parallel across the 2.64-F capacitor. When capacitors are connected in parallel, they share the same voltage.

Therefore, the voltage across the 7.44-F capacitor is also 30.0 V.

So, the voltage across the 7.44-F capacitor is 30.0 V.