Question

Pre-lab EM-5 Ohm's Law and Kirchhoff's Rules Ohm's Law The resistance R of a device can be determined by either directly measuring the resistance using an ohmmeter, or by measuring the current I through it and the voltage Vacross it, and then calculating R using Ohm's Law V R= (1) If the voltage across a resistor is 10V, and the current passing through it is 2.5 mA. of the resistor? What is the resistance R= Ω. Kirchhoffs Loop Rule Around any closed circuit loop, the sum of the potential drops equals the sum of the potential rises Two resistors are connected in series with a 10V battery as shown in Figure 1. If the voltage drop across Ri is V,-3.5 V, what should be the potential drop V2 across R,? R. R, 10V V Figure 1 Kirchhoffs Junction Rule The sum of all currents entering any junction in a circuit is equal to the sum of the currents leaving that junction. Three resistors are connected with a power supply as shown in figure 2. Resistors R2 and Ry are connected (circle one) R, in series. in parallel. R. 1. R, Resistor R, is connected with R2 and R, (circle one) in series. in parallel. Figure 2 If the current passing through resistor R, is 6.00 mA, then what is the sum of the current passing through R and R,? 1 +I, mA. EM-5P.1 November 28, 2017

Answer 1

1) Given values-

V = 10 V

I = 2.5 mA = 2*10^-3 A

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

$R= \frac{10}{(2.5*10^{-3})}$

$R= 4000\Omega$

$R= 4K\Omega$

2) Let, Voltage drop across resistor R₁ is V₁ and voltage drop across R₂ is V₂.

V=10 V

V₁=3.5 V

Using Kirchhoff's Loop Rule,

Sum of voltage difference around a circuit loop is zero.

Therefore,

$V-V_{1}-V_{2}=0$

$V_{2}=V-V_{1}$

$V_{2}= 10-3.5$

$V_{2}= 6.5V$

3)

i) Resistors R₂ and R₃ are connected in parallel.

ii) Resistor R₁ is connected in series with resistors R₂ and R_3.

iii) According to Kirchhoff's junction rule,

$I_{1}= I_{2}+I_{3}$

I₁ = 6 mA (Given)

therefore,

$I_{2}+I_{3}=6 mA$

iv)

R₂ = R₃

therefore,

$I_{2}=I_{3}$ (As both resistors are in parallel, Voltage drop across these resistors are same)     

$I_{1}= I_{2}+I_{3}$

$I_{1}= I_{2}+I_{2}$

$I_{1}= 2I_{2}$

$I_{2}=\frac{I_{1}}{2}$

$I_{2}= 3 mA$

$I_{3}= 3 mA$

v)

As both resistors(R₂ and R₃) are in parallel, Voltage drop across these resistors are same.

$I_{2}*R_{2}= I_{3}*R_{3}$

$\frac{R_{2}}{R_{3}}=\frac{ I_{3}}{I_{2}}$

$\frac{R_{2}}{R_{3}}=\frac{1}{5}$

$\frac{I_{3}}{I_{2}}=\frac{ 1}{5}$

$5I_{3}={I_{2}}$

$I_{1}= I_{2}+I_{3}$

$I_{1}= 5I_{3}+I_{3}$

$I_{1}= 6I_{3}$

$I_{3}= \frac{I_{1}}{6}$

$I_{1}= 6mA$

$I_{3}=1mA$

$I_{2}=5{I_{3}}$

$I_{2}=5mA$

vi)

R₂ and R₃ are in parallel.

therefore equivalent resistance is,

$\frac{1}{R_{23}}=\frac{1}{R_{2}}+\frac{1}{R_{3}}$

$R_{23}=\frac{R_{2}*R_{3}}{R_{2}+R_{3}}$

$R_{23}=\frac{100*500}{100+500}$

$R_{23}=\frac{R_{2}*R_{3}}{R_{2}+R_{3}}$$=\frac{50000}{600}$

$R_{23}=83.33\Omega$

vii)

R₂₃ is equivalent resistance of resistors R₂ and R₃.

R₁ is in series with R₂₃

therefore,

$R_{123}=R_{1}+R_{23}$$=200+83.33$

$R_{123}=283.33$$R_{123}=283.33 \Omega$