Question

The potential difference between the terminals of a battery of emf 6.0 V and internal resistance 1 Ω drops to 5.8 V when connected across an external resistor. Find the resistance of the external resistor.

Answer 1

Answer:

V=E-ir

5.8 = 6 - i(1)

-0.2 = -i

i=0.2 A

v=ir

5.8 = 0.2(r)

r = 29 ohm

Answer 2

The resistance of the external resistor R is 29 Ω .

Explanation:

Step 1:

From the given data, the below information is derived as

Battery voltage (E) = 6 V

Terminal Resistance (R) = 1 Ω

Voltage drop (V) = 5.8 V  

Step 2:

We know that Current  

$\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}=\frac{6}{\mathrm{R}+1}$

$V=E-I r$

$\Rightarrow 5.8=6-\frac{6}{R+1} \times 1$

$\Rightarrow \frac{6}{R+1}=0.2$

$\Rightarrow R+1=30$

$\Rightarrow R=29 \Omega$

⇒Hence resistance (R) = 29 Ω