Two resistances of 400 Omega and 800 Omega ar | vedclass

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

Question

(D) $1$. Potential difference $(V_1)$ across the $400 \Omega$ resistor without the voltmeter:$V_1 = \frac{R_1}{R_1 + R_2} 	imes V = \frac{400}{400 +

Vedclass · Accepted Answer

$0.05$

Vedclass · Answer

(D) $1$. Potential difference $(V_1)$ across the $400 \Omega$ resistor without the voltmeter:$V_1 = \frac{R_1}{R_1 + R_2} 	imes V = \frac{400}{400 + 800} 	imes 6 = \frac{400}{1200} 	imes 6 = 2 	ext{ V}$.$2$. When the voltmeter of resistance $R_v = 10000 \Omega$ is connected in parallel with the $400 \Omega$ resistor,the equivalent resistance $(R_p)$ of this parallel combination is:$R_p = \frac{400 	imes 10000}{400 + 10000} = \frac{4000000}{10400} = \frac{40000}{104} \approx 384.62 \Omega$.$3$. The new potential difference $(V_2)$ across the parallel combination is:$V_2 = \frac{R_p}{R_p + R_2} 	imes V = \frac{384.62}{384.62 + 800} 	imes 6 = \frac{384.62}{1184.62} 	imes 6 \approx 1.948 	ext{ V}$.$4$. The error in measurement is:$	ext{Error} = V_1 - V_2 = 2 - 1.948 = 0.052 	ext{ V}$.Rounding to the nearest given option,the error is approximately $0.05 	ext{ V}$.

Similar Questions

What is the equivalent resistance between $A$ and $B$?

In the given figure-$1$,the resistance of the shown voltmeter is variable. The variation of its reading with respect to its resistance is shown in figure-$2$. The value of $R$ is ............... $\Omega$.

The equivalent resistance between the points $A$ and $B$ will be (each resistance is $15 \Omega$): (in $Omega$)

In the electrical circuit shown in the figure,the current $i$ through the side $AB$ containing the $20 \; \Omega$ resistor is:

The power of an electric motor is $242 \ W$ when connected to a $220 \ V$ supply. When the motor is operated at $200 \ V$,the current drawn by it is (in $A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module