Question

Two resistors, R1 = 2.81Ω and R2 = 5.97Ω , are connected in series to a battery with an EMF of 24.0V and negligible internal resistance. Find the current I1 through R1 and the potential difference V2 across R2 .

Answer 1

Answer:

2.73 A,16.3 V

Explanation:

$R_1=2.81\Omega$

$R_2=5.97\Omega$

$E=24 V$

When internal resistance is negligible

Then, V=E=24 V

In series

$R=R_1+R_2$

Using the formula

$R=2.81+5.97=8.78\Omega$

$V=IR$

$24=I(8.78)$

$I=\frac{24}{8.78}=2.73 A$

In series , current flowing through each resistance remains same and potential across each resistor is different.

Therefore, $I=I_1=I_2=2.73 A$

$V_2=IR_2$

$V_2=2.73\times 5.97=16.3 V$