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.

A) To solve this problem, we need to use the formulas for resistors connected in parallel and power.

(a) The total power supplied by the battery is given as 7.0 W. The power formula is P = V^2 / R, where P is power, V is voltage, and R is resistance.
So, we can write the equation as 7.0 = 30^2 / (R1 + R2), where R1 and R2 are the resistances of the two parallel resistors.
Simplifying the equation, we get 7.0 = 900 / (R1 + R2).

(b) After one resistor is heated, its resistance doubles while the other resistor's resistance remains unchanged.
Let the initial resistance of each resistor be R.
So, the resistance of the heated resistor becomes 2R, and the resistance of the other resistor remains R.
Now, we can use the formula for power to find the total power delivered to the resistors after one resistor has been heated.
The power formula is P = V^2 / R, where P is power, V is voltage, and R is resistance.
So, the power delivered to the heated resistor is P(heated) = 30^2 / (2R), and the power delivered to the other resistor is P(other) = 30^2 / R.
The total power is the sum of these two powers, P(total) = P(heated) + P(other).
Substituting the values, we get P(total) = 30^2 / (2R) + 30^2 / R.

To find the initial resistance of each resistor and the total power after heating, we can solve these equations simultaneously.

B) To find the maximum equivalent capacitance, we need to combine two capacitors in parallel and then connect them in series with the remaining capacitor.

Let's assume C1 and C2 are the two capacitors connected in parallel, and C3 is the remaining capacitor that is connected in series with the parallel combination.

The formula for capacitors connected in parallel is C = C1 + C2, and the formula for capacitors connected in series is 1/C = 1/C1 + 1/C2.

We want to find the maximum equivalent capacitance, which is C.

So, the maximum equivalent capacitance is obtained by connecting C1 and C2 in parallel and then connecting the parallel combination in series with C3.
Therefore, we can write the equation as 1/C = 1/(C1 + C2) + 1/C3.

Substituting the values, we get 1/C = 1/(60 + 48) + 1/36.
Simplifying the equation, we can find the value of 1/C, and then take the reciprocal to find the maximum equivalent capacitance.

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

The 2.64-F and 7.03-F capacitors are connected in series across a 30.0-V battery. The total voltage across them is 30.0 V.

Then, a 7.44-F capacitor is connected in parallel across the 2.64-F capacitor.

When capacitors are connected in series, their charges are the same, but the voltages across them are divided in proportion to their capacitances.

So, we can write the equation as Q1/C1 = Q2/C2, where Q1 is the charge on the 2.64-F capacitor, C1 is its capacitance, Q2 is the charge on the 7.44-F capacitor, and C2 is its capacitance.

Since the voltage across the 2.64-F capacitor is the same as the total voltage of 30.0 V, we can write the equation as Q1/2.64 = 30.0.

Then, when the 7.44-F capacitor is connected in parallel, the total charge is conserved.
So, Q1 = Q2.

By substituting these values, we can calculate the voltage across the 7.44-F capacitor, which is the same as the voltage across the 2.64-F capacitor.