The terminals of battery of emf 12 V and internal resistance 1 Ω are connected to a circular coil of resistance 16 Ω at two points distance a quarter of circumference of coil. The current flowing in smaller arc of circle is

Question

The terminals of battery of emf 12 V and internal resistance 1 Ω are connected to a circular coil of resistance 16 Ω at two points distance a quarter of circumference of coil. The current flowing in smaller arc of circle is (A) 2.4 A (B) 2.25 A (C) 0.75 A (D) 1.66 A

Tardigrade Admin · Accepted Answer

Two parts have resistance of 12 Ω and 4 Ω. Equivalent resistance =(12 × 4/12+4)=3 Ω Current =( text Applied emf / text Equivalent resistance )=(12/4)=3 A <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1d38d471353f1dbbc4e3e69624bbda30-.png" /> Current in R2 resistor, I= Total current × 3 / 4 =2.25 A

Q. The terminals of battery of emf $12 V$ and internal resistance $1 Ω$ are connected to a circular coil of resistance $16 Ω$ at two points distance a quarter of circumference of coil. The current flowing in smaller arc of circle is

2.4 A

2.25 A

0.75 A

1.66 A

Two parts have resistance of $12 Ω$ and $4 Ω$ .
Equivalent resistance $= \frac{12 \times 4}{12 + 4} = 3 Ω$
Current $= \frac{Applied emf}{Equivalent resistance} = \frac{12}{4} = 3 A$

Current in $R_{2}$ resistor, $I =$ Total current $\times 3/4$
$= 2.25 A$

Q. The terminals of battery of emf 12V and internal resistance 1Ω are connected to a circular coil of resistance 16Ω at two points distance a quarter of circumference of coil. The current flowing in smaller arc of circle is