Question

In balanced meter bridge, the resistance of bridge wire is 0.1 Ω cm. Unknown resistance X is connected in left gap and 6 Ω in right gap, null point divides the wire in the ratio 2:3. Find the current drawn from the battery of 5 V having negligible resistance.

पर्याय

• 1 A

• 1.5 A

• 2 A

• 5 A

Answer 1

1 A

Explanation:

For potentiometer,

`"R"_1/"R"_2="l"_1/"l"_2=2/3(∵ "l"_1/"l"_2=2/3)`

`x/6=2/3` ⇒ 4Ωm^-1 

Total resistance = 6 + 4 = 10Ω

Resistance of wire 0.1 × 100 = 10Ω

∴ `1/"R"_"eff"=1/10+1/10`

⇒ `"R"_"eff"` = 5Ω

Current drawn from the battery

l = `"V"/"R"=5/5=1"A"`