$A$ resistor ${R_1}$ dissipates the power $P$ when connected to a certain generator. If the resistor ${R_2}$ is put in series with ${R_1}$,the power dissipated by ${R_1}$

Two resistors of $100 \ \Omega$ and $200 \ \Omega$ are connected in series with a battery of $4 \ V$ and negligible internal resistance. $A$ voltmeter is used to measure the voltage across the $100 \ \Omega$ resistor,which gives a reading of $1 \ V$. The resistance of the voltmeter must be . . . . . . $\Omega$.

The series combination of two batteries,both of the same emf $10 \; V,$ but different internal resistances of $20 \; \Omega$ and $5 \; \Omega,$ is connected to the parallel combination of two resistors $30 \; \Omega$ and $R \; \Omega.$ If the voltage difference across the battery of internal resistance $20 \; \Omega$ is zero,the value of $R$ (in $\Omega$) is:

In the circuit given, $E = 6.0 \, V$, $R_1 = 100 \, \Omega$, $R_2 = R_3 = 50 \, \Omega$, and $R_4 = 75 \, \Omega$. The equivalent resistance of the circuit, in $\Omega$, is:

$A$ straight wire of resistance $R$ is bent into the shape of a square. $A$ cell of emf $12 \text{ V}$ is connected between two adjacent corners of the square. The potential difference across any diagonal of the square is (in $\text{ V}$)

The reading of the voltmeter is :- .............. $V$