The current drawn from a 5 V source in the g | vedclass

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

Question

(B) First,simplify the circuit. The $5 \ \Omega$ and $10 \ \Omega$ resistors are in series,giving $R_1 = 5 + 10 = 15 \ \Omega$.Next,the $10 \ \Omega$

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(B) First,simplify the circuit. The $5 \ \Omega$ and $10 \ \Omega$ resistors are in series,giving $R_1 = 5 + 10 = 15 \ \Omega$.Next,the $10 \ \Omega$ and $20 \ \Omega$ resistors are in series,giving $R_2 = 10 + 20 = 30 \ \Omega$.These two branches ($R_1$ and $R_2$) are in parallel. The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{15} + \frac{1}{30} = \frac{2+1}{30} = \frac{3}{30} = \frac{1}{10}$So,$R_{eq} = 10 \ \Omega$.The current $I$ is given by Ohm's law: $I = \frac{V}{R_{eq}} = \frac{5 \ V}{10 \ \Omega} = 0.5 \ A$.

The current drawn from a $5 \ V$ source in the given circuit is .......... $A$.

Similar Questions

What will be the equivalent resistance of the circuit between $x$ and $y$?

$A$ voltmeter of very high resistance is connected in the circuit as shown in the figure. The voltage shown by this voltmeter will be . . . . . . . (in $V$)

$A$ battery of $e.m.f.$ $10\, V$ is connected to a resistance network as shown in the figure. The potential difference ${V_A} - {V_B}$ between the points $A$ and $B$ is .................... $V$.

When three identical bulbs of $60 \;W, 200 \;V$ rating are connected in series to a $200 \;V$ supply,the power drawn by them will be ....... $Watt$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module