$A$ battery is charged at $15\,V$ for $8\,h$ with a current of $10\,A$. During discharge, the battery supplies a current of $5\,A$ for $15\,h$. The terminal voltage during discharge is $14\,V$. Find the watt-hour efficiency of the battery in percentage.

  • A
    $80$
  • B
    $90$
  • C
    $87.5$
  • D
    $82.5$

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Two voltmeters $V_1$ and $V_2$ are connected in series in a $D.C.$ circuit. The reading of $V_1$ is $80 \, V$ and its sensitivity is $200 \, \Omega/V$. If the resistance of $V_2$ is $32 \, k\Omega$,what is the total voltage of the circuit in $V$?

In Circuit-$1$ and Circuit-$2$ shown in the figures,$R_1=1 \Omega, R_2=2 \Omega$ and $R_3=3 \Omega$. $P_1$ and $P_2$ are the power dissipations in Circuit-$1$ and Circuit-$2$ when the switches $S_1$ and $S_2$ are in open conditions,respectively. $Q_1$ and $Q_2$ are the power dissipations in Circuit-$1$ and Circuit-$2$ when the switches $S_1$ and $S_2$ are in closed conditions,respectively. Which of the following statement$(s)$ is(are) correct?
$(A)$ When a voltage source of $6 V$ is connected across $A$ and $B$ in both circuits,$P_2 > P_1$.
$(B)$ When a constant current source of $2 A$ is connected across $A$ and $B$ in both circuits,$P_1 > P_2$.
$(C)$ When a voltage source of $6 V$ is connected across $A$ and $B$ in Circuit-$1$,$Q_1 > P_1$.
$(D)$ When a constant current source of $2 A$ is connected across $A$ and $B$ in both circuits,$Q_1 > Q_2$.

One junction of a thermocouple is kept at a constant temperature $T_r$ and the other junction is kept at a temperature $T$. If the thermo-emf is given by $E = K(T - T_r) [T_0 - \frac{1}{2}(T + T_r)]$,what is the thermoelectric power at $T = \frac{1}{2}T_0$?

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One junction of a certain thermoelectric couple is at a fixed temperature $T_r$ and the other junction is at temperature $T$. The thermo electromotive force for this is expressed by $E = K(T - T_r) \left[ T_0 - \frac{1}{2}(T + T_r) \right]$. At temperature $T = \frac{1}{2}T_0$,the thermoelectric power is

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When a current $I$ is passed through a wire of constant resistance,it produces a potential difference $V$ across its ends. The graph drawn between $\log I$ and $\log V$ will be:

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