Four resistance of 10 Ω, 60 Ω , 100 Ω and 200 Ω respectively taken in order are used to form a Wheatstone’s bridge. A 15 V battery is connected to the ends of a 200 Ω resistance, the current through it will be

Question

Four resistance of 10 Ω, 60 Ω , 100 Ω and 200 Ω respectively taken in order are used to form a Wheatstone’s bridge. A 15 V battery is connected to the ends of a 200 Ω resistance, the current through it will be (A) 7.5 × 10-5 A  (B) 7.5 × 10-4 A  (C) 7.5 × 10-3 A  (D) 7.5 × 10-2 A

Tardigrade Admin · Accepted Answer

Here, the resistance for 10 Ω, 60 Ω and 100 Ω are in series and they together are in parallel to 200 Ω resistance. When a potential difference of 15 V is applied across 200 Ω then current through it is 1=(15/200)=7.5 × 10-2 A

Q. Four resistance of $10 Ω, 60 Ω, 100 Ω$ and $200 Ω$ respectively taken in order are used to form a Wheatstone’s bridge. A $15 V$ battery is connected to the ends of a $200 Ω$ resistance, the current through it will be

$7.5 \times 1 0^{- 5} A$

$7.5 \times 1 0^{- 4} A$

$7.5 \times 1 0^{- 3} A$

$7.5 \times 1 0^{- 2} A$

Here, the resistance for $10 Ω, 60 Ω$ and $100 Ω$ are in series and they together are in parallel to $200 Ω$ resistance. When a potential difference of $15 V$ is applied across $200 Ω$ then current through it is
$1 = \frac{15}{200} = 7.5 \times 1 0^{- 2} A$

Q. Four resistance of 10Ω,60Ω,100Ω and 200Ω respectively taken in order are used to form a Wheatstone’s bridge. A 15V battery is connected to the ends of a 200Ω resistance, the current through it will be

7.5×10−5A

7.5×10−4A

7.5×10−3A

7.5×10−2A