Q12P Figure shows a resistor of resis... [FREE SOLUTION]

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Chapter 27: Q12P (page 795)

Figure shows a resistor of resistance R= 6.00 鈩� connected to an ideal battery of emf12.0 V by means of two copper wires. Each wire has length 20.0 cm and radius 1.00 mm. In dealing with such circuits in this chapter, we generally neglect the potential differences along the wires and the transfer of energy to thermal energy in them. Check the validity of this neglect for the circuit of Figure: What is the potential difference across (a) The resistor and (b) Each of the two sections of wire? At what rate is energy lost to thermal energy in (c) The resistor And (d) Each section of wire?

Short Answer

Expert verified

Potential difference across the resistor is 12.0V
Potential difference across each of the two sections of wire is $2.15 脳 10^{- 3} V$
The rate at which energy is lost to thermal energy in the resistor is 24W
The rate at which energy is lost to thermal energy in the two sections of wire 4.30mW

Step by step solution

Step 1: Given

$R = 6.00 惟蔚 = 12.0 V L = 20.0 c m = 20.0 脳 10^{- 2} m . r = 1.00 m m = 1.00 脳 10^{- 3} m$

Determining the concept

Find the current resistance of the wire from its resistivity using the corresponding formula. Then, using Ohm鈥檚 law, find the current through the circuit. Using this, easily find the resistance across the wire and the resistor using Ohm鈥檚 law. The formula for the power will give therate at which the energy is lost to the thermal energy in the two sections of the wire and the resistor.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

$R = \frac{蚁 L}{A}$

Where, R is resistance,蚁is resistivity, L is length, A is area.

(a) Determining the potential difference across the resistor

The resistance across each copper wire is,

$R_{w} = \frac{蚁 L}{A}$

$R_{w} = 1.0756 脳 10^{- 3} 惟鈮� 0.0011 惟$

Therefore, the total resistance in the circuit is

$R_{t o t a l} = 2 R_{w} + R R_{t o t a l} = 2 (0.0011) + 6.00 R_{t o t a l} = 6.0022 惟$

Ohm鈥檚 law gives,

V=IR

$I = \frac{V}{R}$

In this case,

$I = \frac{12}{6.0022}$

I=1.9993A

The voltage across the resistor is

V=IR

V=1.9993(6)

=11.996=12V

Hence, potential difference across the resistor is 12.0V.

(b) determining the Potential difference across each of the two sections of wire

The voltage across each wire is,

$V_{w} = I R_{w}$

鈭� V_{w} = 1.9993 (1.0756 脳 10^{- 3})

$鈭� V_{w} = 2.15 脳 10^{- 3} V$

Hence, potential difference across each of the two sections of wire is $鈭� V_{w} = 2.15 脳 10^{- 3} V$

(c) Determining the rate at which energy is lost to thermal energy in the resistor

It is known that,

P=IV

$P = 1.9993 (11.996) P = 23.98 ~ 24 W$

Hence, the rate at which energy is lost to thermal energy in the resistor is 24W

(d) determining the rate at which energy is lost to thermal energy in the two sections of wire

$P_{w} = I V_{w} P_{w} = 1.9993 (2.15 脳 10^{- 3}) P_{w} = 4.298 脳 10^{- 3} = 4.30 m W$

Hence, the rate at which energy is lost to thermal energy in the two sections of wire 4.30mW

Therefore,the potential difference andthe rate at which the energy is lost across the resistor and the wire in the circuit can be found using Ohm鈥檚 law and resistivity of material.