The current I drawn from the 5 text{ V} sourc | vedclass

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

Question

(C) The circuit can be simplified by identifying series and parallel combinations. Looking at the circuit,the $10 \Omega$ and $20 \Omega$ resistors ar

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(C) The circuit can be simplified by identifying series and parallel combinations. Looking at the circuit,the $10 \Omega$ and $20 \Omega$ resistors are in series,giving a total of $30 \Omega$. The $5 \Omega$ and $10 \Omega$ resistors are also in series,giving a total of $15 \Omega$. These two branches ($30 \Omega$ and $15 \Omega$) are in parallel with each other. The equivalent resistance $R_{eq}$ is given by: $R_{eq} = \frac{30 	imes 15}{30 + 15} = \frac{450}{45} = 10 \Omega$. Using Ohm's law,the current $I$ drawn from the $5 	ext{ V}$ source is: $I = \frac{V}{R_{eq}} = \frac{5}{10} = 0.5 	ext{ A}$.

The current $I$ drawn from the $5 \text{ V}$ source will be ............... $A$.

Similar Questions

Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat produced in the two cases is

The equivalent resistance between points $A$ and $B$ in the given network is ............ $\Omega$.

$A$ square-shaped wire with a resistance of $3\, \Omega$ for each side is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in units of $\Omega$ will be:

For what value of $R$ will the net resistance of the circuit be $18 \, \Omega$?

The effective resistance between $A$ and $B$, if the resistance of each resistor is $R$, will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module