Question

A cell of emf 18V has an internal
resistance of 312. The terminal p.d. of the battery becomes 15V when connected by a wire. Find the resistance of the wire. ​

Answer 1

Resistance of the wire = 15 Ω

Given

A cell of emf 18 V has an internal  resistance of 3 Ω.

The terminal p.d. of the battery becomes 15 V ,

when connected by a wire

To Find

Resistance of the wire

Knowledge Required

$\bf \pink{\bigstar\ \; V=\epsilon -ir}$

where ,

• V denotes terminal voltage
• ε denotes emf
• r denotes internal resistance

Solution

• ε = 18 V
• V = 15 V
• r = 3 Ω

Apply formula ,

$\bf \green{\bigstar\ \; V=\epsilon -ir}\\\\\to \rm 15=18-i(3)\\\\\to \rm 3\ i=3\\\\\to \bf i= 1\ A$

Apply Ohms law ,

$\bf \orange{\bigstar\ \; V=iR}\\\\\to \rm 15=(1)R\\\\ \bf \blue{\bigstar\ \; R=15\ \Omega}$

So , Resistance of the wire = 15 Ω

Answer 2

Correct Question:

A cell of EMF 18V has an internal resistance of 3 ohm. The terminal p.d. of the battery becomes 15V when connected by a wire. Find the resistance of the wire.

Explanation:

(Here 'am denoting EMF by e.)

e = V + Ir

Simply substitute the values,

→ 18 = 15 + I(3)

→ 18 - 15 = I(3)

→ 3 = I(3)

→ I = 1A

We have to find the resistance of the wire if the terminal p.d. of the battery becomes 15V.

From ohm's law we can say that,

V = IR

From above data we have I is 1A and given V is 15V. So,

→ 15 = 1(R)

→ 15/1 = R

→ 15 = R

Hence, the resistance of the wire is 15 ohm.