A uniform metallic wire having resistance 4Ω$4\Omega$ is bent to form a square loop (ABCD) (see figure). A resistance of 2Ω$2\Omega$ is connected between points B and D and a battery of 2 V is connected across points A and C as shown in the figure. Now the value of current (I) is
Square loop formed by 4-ohm wire with additional 2-ohm resistor between BD and power supply across AC.
A.2 A
B.1 A
C.3 A
D.4.5 A
Solution & Explanation
Related Formula
I = EReff$$I = \frac{E}{R_{\text{eff}}}$$
Core Logic
Each side of the square loop has resistance 1Ω$1\Omega$.
Square loop formed by 4-ohm wire with additional 2-ohm resistor between BD and power supply across AC.
This constitutes a balanced Wheatstone bridge between points A and C.
No current flows through the central 2Ω$2\Omega$ resistor.
Square loop formed by 4-ohm wire with additional 2-ohm resistor between BD and power supply across AC.
In a metre bridge experiment (see figure), the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer:
Metre bridge setup with cell E and galvanometer G connected.
A. Only the right-sided deflection
B. Only the left-sided deflection
C. There will be no deflection irrespective of the position of the jockey
D. Both right-sided and left-sided deflection and at balance point, no deflection
A resistor is connected to a battery of 12 V emf and internal resistance 2Ω$2\Omega$ . If the current in the circuit is 0.6 A, the terminal voltage of the battery is:
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