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Identical resistances, of value \(R\), each, are connected along the edges of a tetrahedron. If the equivalent resistance of this combination is measured between two vertices, it will be:

1. \(\dfrac{R}{2}\) 2. \(\dfrac{R}{4}\)
3. \(\dfrac{R}{6}\) 4. \(2R\)
Subtopic:  Combination of Resistors |
 62%
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The current \(I_1\) through the \(8~\Omega\) resistance in the figure given below will be:

1. \(1~\text{A}\) 2. \(\dfrac{7}{6} ~\text{A}\)
3. \(\dfrac{5}{6} ~\text{A}\) 4. \(\dfrac{1}{2}~\text{A}\)
Subtopic:  Wheatstone Bridge |
 60%
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A straight wire of resistance \(R\) is shaped into the form of an equilateral triangle and its ends joined. The resistance of this triangle between the two vertices is:

1. \(\dfrac{R}{3}\) 2. \(\dfrac{R}{9}\)
3. \(\dfrac{2R}{9}\) 4. \(\dfrac{2R}{3}\)
Subtopic:  Combination of Resistors |
 64%
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The resistivity of a wire changes gradually, linearly along the length — from \(\rho_0\) to \(2\rho_0\). The total length of the wire is \(l\), while its cross-section is \(A\). The total resistance of the wire is:

1. \(\dfrac{\rho_0 l}{A} \ln 2\) 2. \(\dfrac{\rho_0 l}{2A} \ln 2\)
3. \(\dfrac{3 \rho_{0} l}{A}\) 4. \(\dfrac{3}{2} {\dfrac{\rho_0l}{A}}\)
Subtopic:  Derivation of Ohm's Law |
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A uniformly increasing current flows through a \(30\) \(\Omega\) resistance, as shown in the graph.
             
The thermal energy generated in the resistance due to Joule heating is:
1. \(240\) J
2. \(480\) J
3. \(160\) J
4. \(320\) J
Subtopic:  Heating Effects of Current |
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Two non-ideal batteries are connected in parallel: Battery of EMF \(E_1,\) resistance \(r_1\) and of EMF \(E_2\), resistance \(r_2.\) The resulting equivalent battery has EMF '\(E,\)' resistance \(r.\) If \(r_1<r_2,\) 
1. \(|E-E_1|<|E-E_2|\)
2. \(|E+E_1|<|E+E_2|\)
3. \(|E-E_1|>|E-E_2|\)
4. \(|E+E_1|>|E+E_2|\)
Subtopic:  Grouping of Cells |
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Given below are two statements:
Several resistances \(R_1, R_2, .........R_n\) are connected in parallel. The equivalent resistance of the combination is \(R\).
Assertion (A): The fractional error in \(R\) is most affected by that of the smallest resistance in the combination, other things being equal.
Reason (R): In parallel, the conductances add. The contribution to the overall error in the conductance is largest for the largest conductance or the smallest resistance.
 
1. Both (A) and (R) are True and (R) is the correct explanation of (A).
2. Both (A) and (R) are True but (R) is not the correct explanation of (A).
3. (A) is True but (R) is False.
4. (A) is False but (R) is True.
Subtopic:  Combination of Resistors |
 59%
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In a Wheatstone Bridge arrangement, as shown in the figure, the bridge is balanced. However, when the resistances in the arms \(P,Q\) are switched, the bridge is balanced only when \(R\) is replaced by \(4R\) in the other two arms. If the value of \(R\) is \(100\) \(\Omega\), that of \(S\) is:
1. \(100~\Omega\) 2. \(50~\Omega\)
3. \(200~\Omega\) 4. \(400~\Omega\)
Subtopic:  Wheatstone Bridge |
 56%
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A wire is connected to form an equilateral triangle \(ABC\), each side having a resistance of \(4~\Omega\). The vertex \(C\) is maintained at zero volts (\(V_C=0\)), and currents flowing in at \(A\) and \(B\) are as shown in the figure. The ratio of the potentials at \(D\) and \(E\) \(\Big(i.e.~\frac{V_D}{V_E}\Big)\) equals:
1. \(\dfrac31\) 2. \(\dfrac21\)
3. \(\dfrac11\) 4. \(\dfrac53\)
Subtopic:  Kirchoff's Current Law |
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\(AB\) is a \(20~\Omega\) resistor with a tapping point \(C\) that can be moved along \(AB\). The resistances in \(AC,BC\) are proportional to the lengths \(AC,BC\). Initially, \(C\) is at the mid-point of \(AB\) and the circuit is switched on.
                           
If the tapping point \(C\) is moved so that the length \(BC\) is reduced to half its initial value, then the voltage across the \(15~\Omega\) resistor,
1. increases by \(1\) V
2. decreases by \(1\) V
3. increases by \(3\) V
4. decreases by \(3\) V
Subtopic:  Kirchoff's Voltage Law |
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