Two batteries $A$ and $B$ each of $e.m.f.$ $2\, V$ are connected in series to an external resistance $R = 1 \,ohm$. If the internal resistance of battery $A$ is $1.9\, ohms$ and that of $B$ is $0.9\, ohm$, what is the potential difference between the terminals of battery $A$ ............. $V$
For cell $A$ $E = V + ir$ $ \Rightarrow $ $V = 2 - \frac{4}{{3.8}} \times 1.9 = 0$.
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.*
For the arrangement of the potentiometer shown in the figure, the balance point is obtained at $a$ distance $75\,cm $ from $A$ when the key $k$ is open. The second balance point is obtained at $60\, cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $ \Omega$) of the battery $E_1$.
A wire has a non-uniform cross-section as shown in figure.A steady current flows through it. The drift speed of electrons at points $P$ and $q$ is $v_P$ and $v_Q$.
When the resistance $R$ (indicated in the figure below) is changed from $1 \,k \Omega$. to $10 \,k \Omega$, the current flowing through the resistance $R'$ does not change. What is the value of the resistor $R'?$
A potentiometer consists of a wire of length $4\, m$ and resistance $10\,\Omega $. It is connected to a cell of $e.m.f.$ $2\, V$. The potential difference per unit length of the wire will be ............. $V/m$