Question

Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ohm are joined in series to provide a supply to a resistance of 8.5 Ohm . What is the current drawn from the supply and its terminal voltage?

Answer 1

Given Data:

• Number of cells, n = 6
• EMF of each cell, E_cell = 2.0 V
• Internal resistance of each cell, r_cell = 0.015 Ω
• External resistance (load), R = 8.5 Ω

Step 1: Calculate the Total EMF and Total Internal Resistance of the Battery

Since the six cells are connected in series, we find the total EMF and total internal resistance by summing the values for each cell.

Total Electromotive Force (EMF):
The total EMF (Etotal) is the sum of the individual EMFs.
$E{total} = n \times E{cell}$
$E{total} = 6 \times 2.0 \ V = 12.0 \ V$

Total Internal Resistance:
The total internal resistance (rtotal) is the sum of the individual internal resistances.
$r{total} = n \times r{cell}$
$r{total} = 6 \times 0.015 \ Ω = 0.09 \ Ω$

Step 2: Calculate the Current Drawn from the Supply

Now we can treat the battery as a single source with an EMF of 12.0 V and an internal resistance of 0.09 Ω.

Using Ohm's Law for the entire circuit, the current (I) is the total EMF divided by the total resistance (external + internal).

$I = \frac{E{total}}{R + r{total}}$

$I = \frac{12.0 \ V}{8.5 \ Ω + 0.09 \ Ω}$

$I = \frac{12.0 \ V}{8.59 \ Ω}$

$I \approx 1.397 \ A$

So, the current drawn from the supply is approximately 1.40 A.

Step 3: Calculate the Terminal Voltage

The terminal voltage (V) is the potential difference across the terminals of the battery when it is supplying current. It can be calculated in two ways:

Method 1: Voltage drop across the external resistor
The terminal voltage is equal to the voltage across the external resistance R.
$V = I \times R$
$V = 1.397 \ A \times 8.5 \ Ω$
$V \approx 11.875 \ V$

Method 2: EMF minus the voltage drop across the internal resistance
The terminal voltage is the total EMF minus the voltage "lost" due to the internal resistance.
$V = E{total} - I \times r{total}$
$V = 12.0 \ V - (1.397 \ A \times 0.09 \ Ω)$
$V = 12.0 \ V - 0.1257 \ V$
$V \approx 11.874 \ V$

Both methods give the same result (the small difference is due to rounding the current).

Rounding to an appropriate number of significant figures, the terminal voltage is 11.9 V.

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Final Answer:

• The current drawn from the supply is 1.40 A.
• The terminal voltage of the supply is 11.9 V.