Four resistances $10$ $\Omega$, $5$ $\Omega$, $7$ $\Omega$ and $3$ $\Omega$ are connected so that they form the sides of a rectangle $AB$, $BC$, $CD$ and $DA$ respectively. Another resistance of $10$ $\Omega$ is connected across the diagonal $AC$. The equivalent resistance between $A$ and $B $ is .............. $\Omega$
A$2$
B$5$
C$7$
D$10$
Medium
Download our app for free and get started
B$5$
b The figure can be drawn as follows
$ \Rightarrow $ ${R_{AB}} = 5\,\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.*
In a Wheatstone’s bridge all the four arms have equal resistance $R$. If the resistance of the galvanometer arm is also $R$, the equivalent resistance of the combination as seen by the battery is
In the adjoining circuit, the battery $E_1$ has an emf of $12\, volt$ and zero internal resistance while the battery $E_2$ has an $emf$ of $2\, volt$. If the galvanometer $G$ reads zero, then the value of the resistance $X$ (in $ohm$ ) is
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
In the diagram shown, the reading of voltmeter is $20\, V$ and that of ammeter is $4\, A$. The value of $R$ should be (Consider given ammeter and voltmeter are not ideal)