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Chapter 14: Problem 70

In the given circuit, find the equivalent resistance between points \(A\) and \(B\). (A) \(18 \Omega\) (B) \(12 \Omega\) (C) \(20 \Omega\) (D) \(27 \Omega\)

Short Answer

Expert verified
In this case, a valid short answer cannot be generated without a circuit diagram provided. Please provide a circuit diagram to obtain the correct equivalent resistance.

Step by step solution

01

Identify the circuit connections

Observe the given circuit and identify whether resistances are connected in series, in parallel, or a combination of both. Unfortunately, the given problem does not provide an image of the circuit, so we will assume the circuit has a combination of series and parallel connections. Without a visual representation of the circuit, we cannot proceed with breaking down the connections and solving for the equivalent resistance. If the exercise includes an image, please provide it for a more accurate solution.

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Most popular questions from this chapter

A wire of resistance \(4 \Omega\) is stretched to twice its original length. What is the resistance of the wire now? (A) \(1 \Omega\) (B) \(14 \Omega\) (C) \(8 \Omega\) (D) \(16 \Omega\)

A wire \(l=8 \mathrm{~m}\) long of uniform cross-sectional area \(A=8 \mathrm{~mm}^{2}\) has a conductance of \(G=2.45 \Omega^{-1}\). The resistivity of material of the wire will be (A) \(2.1 \times 10^{-7} \Omega \mathrm{m}\) (B) \(3.1 \times 10^{-7} \Omega \mathrm{m}\) (C) \(4.1 \times 10^{-7} \Omega \mathrm{m}\) (D) \(5.1 \times 10^{-7} \Omega \mathrm{m}\)

A galvanometer having a coil resistance of \(100 \Omega\) gives a full scale deflection, when a current of \(1 \mathrm{~mA}\) is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of \(10 \mathrm{~A}\), is (A) \(2 \Omega\) (B) \(0.1 \Omega\) (C) \(3 \Omega\) (D) \(0.01 \Omega\)

The current at which a fuse wire melts does not depend on its (A) Cross-sectional area (B) Length (C) Resistivity (D) Density

The resistance of a wire is \(10 \Omega\). Its length is increased by \(10 \%\) by stretching. The new resistance will now be nearly (A) \(12 \Omega\) (B) \(1.2 \Omega\) (C) \(13 \Omega\) (D) \(11 \Omega\)
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