The terminal voltage of the battery, whose emf is\(10V\) and internal resistance\(1Ω\), when connected through an external resistance of \(4Ω\) as shown in the figure is:
The terminal voltage is given by:
$V_{terminal} = E - Ir$
where $E$ is the emf, $I$ is the current, and $r$ is the internal resistance.
Total resistance, $R_\text{total} = 4 \, \Omega + 1 \, \Omega = 5 \, \Omega$.
Current, $I = \frac{E}{R_\text{total}} = \frac{10}{5} = 2 \, A$.
Terminal voltage, $V_{terminal} = 10 - 2 \times 1 = 8 \, V$.
The terminal voltage (\(V\)) of a battery is given by:
V = E − I × r
Where \(E = 10V\) is the emf, \(r = 1\Omega\) is the internal resistance, and \(I\) 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 \(100\) is divided into 10 equal parts. The first \(5\) parts are connected in series while the next \(5\) 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 | Down’s syndrome | I | 11th chormosome |
B | α-Thalassemia | II | ‘X’ chromosome |
C | β-Thalassemia | III | 21st chromosome |
D | Klinefelter’s syndrome | IV | 16th chromosome |
The velocity (v) - time (t) plot of the motion of a body is shown below :
The acceleration (a) - time(t) graph that best suits this motion is :