a series of six lead accumulators,each of emf 2.0 V and internal resistance 0.50 ohm is charged by a 100 V dc supply. (a) What series resistance should be used in the charging circuit in order to limit the current to 8.0Ampere? (b) using the required resistance,obtain: (i) the power supplied by dc source, (ii)the power dissipated as heat,and (iii) the energy stored in the battery in 15 min.

To tackle this problem, we need to break it down into manageable parts. We have a series of six lead accumulators, each with an electromotive force (emf) of 2.0 V and an internal resistance of 0.50 ohm. The charging is done using a 100 V DC supply, and we want to limit the current to 8.0 A. Let's go through the calculations step by step.

Determining the Required Series Resistance

First, we need to find the total voltage of the six lead accumulators. Since they are in series, we can simply multiply the emf of one accumulator by the number of accumulators:

• Total emf = 6 accumulators × 2.0 V = 12.0 V

Next, we calculate the total internal resistance of the six accumulators. Again, since they are in series, we add their internal resistances:

• Total internal resistance = 6 accumulators × 0.50 ohm = 3.0 ohm

Now, we can apply Ohm's Law and the formula for the total voltage in the circuit. The total voltage from the DC supply is 100 V, and we want the current to be limited to 8.0 A. The total resistance in the circuit (R_total) can be calculated using the formula:

• R_total = V_supply / I = 100 V / 8.0 A = 12.5 ohm

Since we already have the internal resistance of the accumulators (3.0 ohm), we can find the required series resistance (R_series) by subtracting the internal resistance from the total resistance:

• R_series = R_total - R_internal = 12.5 ohm - 3.0 ohm = 9.5 ohm

Calculating Power and Energy

Now that we have the required series resistance, we can move on to the next part of the question. We need to calculate:

(i) Power Supplied by the DC Source

The power supplied by the DC source (P_supply) can be calculated using the formula:

• P_supply = V_supply × I = 100 V × 8.0 A = 800 W

(ii) Power Dissipated as Heat

The power dissipated as heat in the series resistance can be calculated using the formula:

• P_heat = I² × R_series = (8.0 A)² × 9.5 ohm = 64 A² × 9.5 ohm = 608 W

(iii) Energy Stored in the Battery

To find the energy stored in the battery over a period of 15 minutes, we first need to calculate the total charge (Q) that flows into the battery during this time. The formula for charge is:

• Q = I × t

Where:

• I = 8.0 A
• t = 15 minutes = 15 × 60 seconds = 900 seconds

Now, substituting the values:

• Q = 8.0 A × 900 s = 7200 C

The energy stored in the battery (E) can be calculated using the formula:

• E = Q × V_battery

Where V_battery is the total emf of the batteries (12.0 V):

• E = 7200 C × 12.0 V = 86400 J

Summary of Results

To summarize:

• Required series resistance: 9.5 ohm
• Power supplied by the DC source: 800 W
• Power dissipated as heat: 608 W
• Energy stored in the battery in 15 minutes: 86400 J

This breakdown should help you understand how to approach similar problems involving circuits, resistances, and energy calculations. If you have any further questions or need clarification on any of these steps, feel free to ask!