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Chapter 12: Problem 1791

Find the equivalent resistance a cross \(\mathrm{AB}\) (A) \(1 \Omega\) (B) \(2 \Omega\) (C) \(3 \Omega\) (D) \(4 \Omega\)

Short Answer

Expert verified
The given exercise does not provide enough information about the values of the resistors or their connections in the circuit. Please provide a circuit diagram or more information about the circuit components to find the equivalent resistance across AB.

Step by step solution

01

Identify Circuit Connections

First, we need to identify the connections between resistors in the circuit. In this case, there are no clarifications about the components present in the circuit. We will assume there are resistors with equal values, connected either in series or parallel.
02

Series Connection Formula

The total resistance \(R_s\) for resistors connected in series can be calculated using the series connection formula: \[ R_s = R_1 + R_2 + R_3 + ... + R_n \] where \(R_1, R_2, R_3, ..., R_n\) are the values of resistors connected in series.
03

Parallel Connection Formula

The total resistance \(R_p\) for resistors connected in parallel can be calculated using the parallel connection formula: \[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n} \] where \(R_1, R_2, R_3, ..., R_n\) are the values of resistors connected in parallel.
04

Calculate Equivalent Resistance

To find the equivalent resistance across AB, we will use the series and parallel connection formulas as needed for the given circuit configuration. Unfortunately, the given exercise does not provide enough information about the values of the resistors or their connections in the circuit. Please provide a circuit diagram or more information about the circuit components to find the equivalent resistance across AB.

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Most popular questions from this chapter

n identical cells each of e.m.f \(E\) and internal resistance \(r\) are connected in series An external resistance \(R\) is connected in series to this combination the current through \(\mathrm{R}\) is. (A) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{nr})\\}\) (B) \(\\{(\mathrm{nE}) /(\mathrm{n} R+\mathrm{r})\\}\) (C) \(\\{\mathrm{E} /(\mathrm{R}+\mathrm{nr})\\}\) (D) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{r})\\}\)

The network is made of uniform wire. The resistance of portion EL is $2 \Omega\(. Find the resistance of star between points \)F \& C .$ (A) \(0.985 \Omega\) (B) \(1.25 \Omega\) (C) \(1.946 \Omega\) (D) \(1.485 \Omega\)

The tungsten filament of bulb has resistance equal to \(18 \Omega\) at \(27^{\circ} \mathrm{C}\) temperature \(0.25 \mathrm{~A}\) of current flows, when \(45 \mathrm{~V}\) is connected to it If $\alpha=4.5 \times 10^{-3} \mathrm{~K}^{-1}$ for a tungsten then find the temperature of the filament. (A) \(2160 \mathrm{~K}\) (B) \(1800 \mathrm{~K}\) (C) \(2070 \mathrm{~K}\) (D) \(2300 \mathrm{~K}\)

The resistivity of a potentiometer wire is $40 \times 10^{-8} \mathrm{ohm} \mathrm{m}\( and its area of cross-section is \)8 \times 10^{-6} \mathrm{~m}^{2}\(. If \)0.2$ amp current is flowing through the wire, the potential gradient will be. (A) \(10^{-2}\) Volt \(/ \mathrm{m}\) (B) \(10^{-1}\) Volt \(/ \mathrm{m}\) (C) \(3.2 \times 10^{-2}\) Volt \(/ \mathrm{m}\) (D) 1 Volt \(/ \mathrm{m}\)

Two heater wires of equal length are first connected in series and then in parallel The ratio of heat produced in the two cases is (A) \(2: 1\) (B) \(1: 2\) (C) \(4: 1\) (D) \(1: 4\)
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