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
A$10$
B$100$
C$500$
D$200$
Medium
Download our app for free and get started
B$100$
b Current across $(500 \Omega) \rightarrow$
$I=\frac{E_{1}-E_{2}}{500}$ ............$(1)$
$\mathrm{E}_{2}=\mathrm{IX}$ .........$(2)$
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.*
A potentiometer wire is $100\,\, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50\,\, cm$ and $10\,\, cm$ from the positive end of the wire in the two cases. The ratio of emf's is
A student measures the terminal potential difference $(V)$ of a cell $($ of $emf \;\varepsilon $ and internal resistance $r)$ as a function of the current $(I)$ flowing through it. The slope, and intercept, of the graph between $V$ and $I$, then, respectively, equal
The circuit diagram shown consists of a large number of element (each element has two resistors $R_1$ and $R_2$). The resistance of the resistors in each subsequent element differs by $a$ factor of $K = 1/2$ from the resistance of the resistors in the previous elements. The equivalent reistance between $A$ and $B$ shown in figure is :
In the circuit shown below the resistance of the galvanometer is $20\, \Omega$ . In which case of the following alternatives are the currents arranged strictly in the decreasing order