Practice free Current Electricity (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The magnitude and direction of the current in the following circuit is:
Net EMF $= 10 - 5 = 5$ V (cells oppose), total $R = 2+1+7 = 10\ \Omega$, current $= 0.5$ A driven by the 10 V cell, so from $A$ to $B$ through $E$.
The resistance of platinum wire at $0^\circ$C is $2\ \Omega$ and $6.8\ \Omega$ at $80^\circ$C. The temperature coefficient of resistance of the wire is:
$R_t = R_0(1+\alpha t)\Rightarrow 6.8 = 2(1+80\alpha)\Rightarrow \alpha = 0.03 = 3\times 10^{-2}\ ^\circ\text{C}^{-1}$.
10 resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is:
Series: $I_s = E/(10R)$. Parallel: $I_p = E/(R/10) = 10E/R$. $n = I_p/I_s = 100$.
A wire of length '$l$' and resistance $100\ \Omega$ 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:
Each part $= 10\ \Omega$; series of 5 = 50; parallel of 5 = 2; total = 52.
The terminal voltage of the battery, whose emf is $10V$ and internal resistance $1\ \Omega$, when connected through an external resistance of $4\ \Omega$ as shown in the figure is:
$I = 10/5 = 2$ A; $V_\text{term} = 2 \times 4 = 8$ V.
Choose the correct circuit which can achieve the bridge balance. (Each option shows a four-arm Wheatstone arrangement with resistors 10 Ω, 10 Ω, 15 Ω and 5 Ω, a galvanometer $G$, a key $K$ and a cell $E$.)
A balanced Wheatstone requires $P/Q = R/S$ with $G$ on the bridge diagonal and the cell on the other diagonal.
Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
$R_A = 2R_B$; $P_\text{series}/P_\text{parallel} = (V^2/3R_B)/(3V^2/2R_B) = 2/9$.
The current passing through the battery in the given circuit, is:
The 6 Ω arm bridges the balanced Wheatstone network, so it carries no current. The remaining resistors reduce to an effective resistance giving $I = \dfrac{5\ \text{V}}{2.5\ \Omega} = 2.0\ \text{A}$ through the battery.
A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
Each piece $= R/8$. Four in parallel $= \dfrac{R/8}{4} = \dfrac{R}{32}$. Two such sets in series $= \dfrac{R}{32}+\dfrac{R}{32} = \dfrac{R}{16}$.
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:
With CD a connecting wire, $V_M = 32$ V. Current in $1\,\Omega$ (A→C) $= 18$ A and in $2\,\Omega$ (C→B) $= 16$ A, so $I_{CD} = 18-16 = 2.0$ A.
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