JEE Physics Current Electricity MCQs Set 09

Answer: (b) 3.2 volt

Explanation:
1. The 4 ohm and 3 ohm resistors are in parallel, so the voltage across the upper branch is \(4\times1=4\) V
2. The lower branch is across the same 4 V. Its resistance is \((0.5\parallel0.5)+1=0.25+1=1.25\) ohm, so its current is \(4/1.25=3.2\) A
3. \(V_{MN}\) is the drop across the 1 ohm resistor: \(3.2\times1=3.2\) V

Teacher's Note:
1. Branches connected between the same two terminals always have the same voltage. Find the branch current first and then the drop across the required resistor.

Answer: (a) current

Explanation:
1. The conductor behaves like resistors in series, so charge cannot accumulate and the current is the same at every section
2. Current density, drift velocity and field all depend on the cross-section \(A\) (\(J=I/A\)) and so change along the conductor

Teacher's Note:
1. Current is a scalar flow that stays the same in a series path; the other quantities vary with area.

Answer: (d) \(V_A=+1.5\ \text{V},\ V_D=-0.5\ \text{V}\)

Explanation:
1. Current flows from high to low potential, so \(V_A=V_B+1\times1.5=1.5\) V
2. \(V_C=V_B-1\times2.5=-2.5\) V
3. Going through the cell from its negative to its positive terminal raises the potential by 2 V: \(V_D=-2.5+2=-0.5\) V

Teacher's Note:
1. Potential falls along the current through a resistor and rises when you cross a cell from the negative to the positive terminal.

Answer: (d) 0 A

Explanation:
1. Take the junction between the two cells as reference. Each cell has its positive terminal on the outer point, so \(V_A-V_M=5\) V and \(V_C-V_M=5\) V
2. Hence \(V_A=V_C\)
3. No potential difference exists across the 12 ohm resistor, so no current flows through it

Teacher's Note:
1. Two points joined by cells that raise them by the same amount from a common junction are at the same potential. A resistor between equipotential points carries no current.

Answer: (c) \(\frac11\)

Explanation:
1. For a square plate of side \(l\) and thickness \(t\): \(R=\rho\frac{l}{l\,t}=\frac{\rho}{t}\)
2. The side \(l\) cancels, so the resistance depends only on the material and thickness
3. Both plates have the same \(\rho\) and \(t\), so \(R_A/R_B=1\)

Teacher's Note:
1. The resistance of a square sheet is independent of its size. This is the idea of sheet resistance used for thin films.

Answer: (c) \(\frac R4\)

Explanation:
1. The ring has no resistance, so all four outer ends of the resistors are at the potential of \(b\)
2. The four resistors are therefore in parallel between \(a\) and \(b\)
3. \(R_{eq}=R/4\)

Teacher's Note:
1. A wire of zero resistance joins all points on it to one node. Then identical resistors become a simple parallel combination.

Answer: (d) 0.5

Explanation:
1. Key open: only one 40 ohm resistor, \(i_1=12/40=0.3\) A
2. Key closed: two 40 ohm resistors in parallel give 20 ohm, \(i_2=12/20=0.6\) A
3. \(i_1/i_2=0.3/0.6=0.5\)

Teacher's Note:
1. Closing the key adds a parallel path, which halves the resistance and doubles the current.

Answer: (b) 1.5 V

Explanation:
1. \(E=I(R+r)\) gives \(0.5(2+r)=0.25(5+r)\)
2. \(1+0.5r=1.25+0.25r\), so \(r=1\) ohm
3. \(E=0.5\times(2+1)=1.5\) V

Teacher's Note:
1. Write the cell equation twice and solve for the internal resistance first.

Answer: (d) 0.5 A

Explanation:
1. Two 2 ohm resistors in parallel give 1 ohm
2. Total resistance = internal 1 + 2 + 1 + 4 = 8 ohm (ideal ammeter has no resistance)
3. \(I=4/8=0.5\) A

Teacher's Note:
1. An ideal ammeter adds no resistance. Reduce the parallel part first and then add all series resistances including the internal resistance.

Answer: (a) 2 V

Explanation:
1. Each of the two parallel branches has resistance 6 ohm, so the combination is 3 ohm; with the 2 ohm resistor the total is 5 ohm and \(I=10/5=2\) A
2. Each branch carries 1 A. The current flows through the branches from right to left (positive terminal of the cell is on the right)
3. \(V_A=V_{right}-1\times2\) and \(V_B=V_{right}-1\times4\)
4. \(V_A-V_B=-2+4=2\) V

Teacher's Note:
1. Potential falls in the direction of the current. Compare the drops from the common right-hand node to A and to B.