In the circuit given $E = 6.0 \,V, R_1 = 100\, ohms, R_2 = R_3 = 50\, ohms, R_4 = 75\, ohms$. The equivalent resistance of the circuit, in $ohms$, is
A$11.875$
B$26.31$
C$118.75$
D
None of these
Medium
Download our app for free and get started
C$118.75$
c (c) In given circuit three resistance ${R_2},{R_4}$ and ${R_3}$ are parallel.
$\frac{1}{R} = \frac{1}{{{R_2}}} + \frac{1}{{{R_4}}} + \frac{1}{{{R_3}}}$
$ = \frac{1}{{50}} + \frac{1}{{50}} + \frac{1}{{75}}$
$ = \frac{{75 + 75 + 50}}{{50 \times 75}}$
$R = \frac{{50 \times 75}}{{75 + 75 + 50}} = \frac{{50 \times 75}}{{200}} = \frac{{75}}{4}\,\Omega = 18.75\,\Omega $
This resistance is in series with ${R_1}$
${R_{{\rm{resultant}}}} = {R_1} + R = 100 + 18.75 = 118.75\,\Omega $
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance $2000 \,\Omega$ is used to measure the potential difference across $500 \,\Omega$ resister, the reading of the voltmeter will be ............... $V$
The sliding contact $C$ is at one fourth of the length of the potentiometer wire $( AB )$ from $A$ as shown in the circuit diagram. If the resistance of the wire $AB$ is $R _0$, then the potential drop $( V )$ across the resistor $R$ is