In the circuit shown,the currents i_{1} and i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) First,calculate the equivalent resistance of the parallel combination of $12 \Omega$ and $4 \Omega$ resistors:$R_p = \frac{12 	imes 4}{12 + 4} =

Vedclass · Accepted Answer

$i_{1}=0.5 	ext{ A}, i_{2}=1.5 	ext{ A}$

Vedclass · Answer

(B) First,calculate the equivalent resistance of the parallel combination of $12 \Omega$ and $4 \Omega$ resistors:$R_p = \frac{12 	imes 4}{12 + 4} = \frac{48}{16} = 3 \Omega$Now,calculate the total resistance of the circuit including the series resistor and internal resistance:$R_{total} = R_p + 2 \Omega + 1 \Omega = 3 \Omega + 2 \Omega + 1 \Omega = 6 \Omega$The total current $I$ flowing from the battery is:$I = \frac{V}{R_{total}} = \frac{12 	ext{ V}}{6 \Omega} = 2 	ext{ A}$Using the current divider rule for parallel resistors:$i_1 = I 	imes \frac{R_2}{R_1 + R_2} = 2 	imes \frac{4}{12 + 4} = 2 	imes \frac{4}{16} = 0.5 	ext{ A}$$i_2 = I 	imes \frac{R_1}{R_1 + R_2} = 2 	imes \frac{12}{12 + 4} = 2 	imes \frac{12}{16} = 1.5 	ext{ A}$Thus,$i_1 = 0.5 	ext{ A}$ and $i_2 = 1.5 	ext{ A}$.

In the circuit shown,the currents $i_{1}$ and $i_{2}$ are

Similar Questions

The reading of the ammeter as per the figure shown is:

For the given circuit,the $emf$ of the cells are $E_1 = 4 \, V$ and $E_2 = 8 \, V$,and their internal resistances are $0.5 \, \Omega$ and $1 \, \Omega$ respectively. Calculate the terminal voltage of each cell.

In the circuit shown,the galvanometer $G$ of resistance $60 \, \Omega$ is shunted by a resistance $r = 0.02 \, \Omega$. The current through $R$ is nearly $1 \, A$. The value of resistance $R$ (in $\Omega$) is nearly:

The charge flowing through a resistor $R$ varies with time $t$ as $Q = 3t - 6t^2$. The heat produced in $R$ until the current in it becomes zero is:

The resistance of an ideal voltmeter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module