A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3V and a resistance of 10 Ω . The potential gradient along the wire in V/m is

Question

A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3V and a resistance of 10 Ω . The potential gradient along the wire in V/m is (A) 0.02 (B) 0.1 (C) 0.2 (D) 1.2

Tardigrade Admin · Accepted Answer

As resistances are in series Rtotal=R1+R2 =20+10=30Ω i=(4/R texttotal)=(3/30)=(1/10) textA So V textwire= iR textwire ⇒ V textwire=(1/10)× 20=2V Hence, potential gradient is (V textwire/l)=(2/10)=0.2 textV/m

Q. A 10 m long wire of resistance 20 $Ω$ is connected in series with a battery of emf 3V and a resistance of 10 $Ω$ . The potential gradient along the wire in V/m is

0.02

0.1

0.2

1.2

As resistances are in series $R_{t o t a l} = R_{1} + R_{2}$ $= 20 + 10 = 30Ω$ $i = \frac{4}{R _{total}} = \frac{3}{30} = \frac{1}{10} A$ So $V_{wire} = i R_{wire}$ $\Rightarrow$ $V_{wire} = \frac{1}{10} \times 20 = 2 V$ Hence, potential gradient is $\frac{V _{wire}}{l} = \frac{2}{10} = 0.2 V/m$

Q. A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3V and a resistance of 10 Ω . The potential gradient along the wire in V/m is

0.02

0.1

0.2

1.2

As resistances are in series Rtotal​=R1​+R2​ =20+10=30Ω i=Rtotal​4​=303​=101​A So Vwire​=iRwire​ ⇒ Vwire​=101​×20=2V Hence, potential gradient is lVwire​​=102​=0.2V/m

Q. A 10 m long wire of resistance 20 $Ω$ is connected in series with a battery of emf 3V and a resistance of 10 $Ω$ . The potential gradient along the wire in V/m is

As resistances are in series $R_{t o t a l} = R_{1} + R_{2}$ $= 20 + 10 = 30Ω$ $i = \frac{4}{R _{total}} = \frac{3}{30} = \frac{1}{10} A$ So $V_{wire} = i R_{wire}$ $\Rightarrow$ $V_{wire} = \frac{1}{10} \times 20 = 2 V$ Hence, potential gradient is $\frac{V _{wire}}{l} = \frac{2}{10} = 0.2 V/m$