The terminal voltage of the battery, whose emf is and internal resistance, when connected through an external resistance of as shown in the figure is:
The terminal voltage is given by:
where is the emf, is the current, and is the internal resistance.
Total resistance, .
Current, .
Terminal voltage, .
The terminal voltage () of a battery is given by:
V = E − I × r
Where is the emf, is the internal resistance, and is the current.
Using Ohm’s law, the total resistance in the circuit is:
Rtotal = Rext + r = 4Ω + 1Ω = 5Ω
The current is:
I = E / Rtotal = 10V / 5Ω = 2A
Now, substitute the values to find the terminal voltage:
V = E − I × r = 10V − (2 × 1Ω) = 8V
A wire of length and resistance is divided into 10 equal parts. The first parts are connected in series while the next parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
List-I | List-II | ||
(A) | 1 mol of H2O to O2 | (I) | 3F |
(B) | 1 mol of MnO-4 to Mn2+ | (II) | 2F |
(C) | 1.5 mol of Ca from molten CaCl2 | (III) | 1F |
(D) | 1 mol of FeO to Fe2O3 | (IV) | 5F |
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The potential (V) at any axial point, at 2 m distance(r) from the centre of the dipole of dipole moment vector
of magnitude, 4 × 10-6 C m, is ± 9 × 103 V.
(Take SI units)
Reason R : , where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below :