Question

Repeat the calculations in Problem 2-11, assuming that V_B = 12.5 V and R = 0.40 k$\Omega$

Short answer

The percentage change in current is -4.4%

Step-by-step solution

Given information: 

The following diagram is given:

Here V_B= 12.5V and R = 0.4k$\Omega$.

The resistance of the cell changes from 25$\Omega$to 45$\Omega$

Calculations:

Initially, the resistance of the cell is 25 Ohms.

Let R_c represents the resistance of the cell.

So initial current is:

i = V_B/ (R+R_C)

i = 12.5 / (400+25)

i = 29.4 mA

Finally the resistance of the cell is 45 Ohms. So the final current is:

I = V_B/ (R+R_C)

I = 12.5 / (400+45)

I = 28.1 mA

Change in current:

I - i = -1.3 mA

Percentage change:

= {-1.3 / 29.4} x 100%

= -4.4%