A cell of emf 2v and internal resistance of 1ohms passes a current through an external load of 9ohms.calculate the potential difference across the cell

Answer 1

1/10 of the voltage will be lost in internal resistance, so output voltage will be 2*9/10 Volts=1.8volts

Now the long way.
current=2v/10ohms=.2amps
voltage lost in internal resistance: .2*1=.2volts
terminal voltage=2-.2=1.8volts

Answer 2

Good

Answer 3

a cell has an e.m.f of 3v when it is connected across a resistor of resistance 4 a current of 0.5 a passes through the circuit. Calculate the internal resistance of the cell.

Answer 4

0.2v

Answer 5

To calculate the internal resistance of the cell, we need to use Ohm's law:

V = IR

where V is the voltage across the resistor, I is the current passing through the circuit, and R is the resistance of the resistor.

In this case, V = 3V, I = 0.5A, and R = 4Ω. Therefore:

3V = 0.5A x 4Ω + V_internal

Simplifying:

3V = 2V + V_internal

V_internal = 1V

Now we can use another formula:

V_internal = I_internal x R_internal

where I_internal is the current passing through the internal resistance of the cell, and R_internal is the internal resistance itself.

We can rearrange this to get:

R_internal = V_internal / I_internal

We don't know the value of I_internal, but we can use Kirchhoff's circuit laws to find it. The current passing through the external resistor is the same as the current passing through the internal resistance and the cell. Therefore:

I_total = I_internal + I_external

I_total = 0.5A

I_external = 0.5A

Therefore:

I_internal = I_total - I_external = 0A

This means that there is no current passing through the internal resistance of the cell. Therefore, the internal resistance is infinite.

In summary, the internal resistance of the cell is infinite.