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Q54PE

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Found in: Page 777

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A ${\mathbf{0}}{\mathbf{.}}{\mathbf{0200}}{\mathbf{}}{\mathbf{\Omega }}$ ammeter is placed in series with a ${\mathbf{10}}{\mathbf{.}}{\mathbf{00}}{\mathbf{‐}}{\mathbf{\Omega }}$ resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) Calculate the resistance of the combination. (c) If the voltage is kept the same across the combination as it was through the ${\mathbf{10}}{\mathbf{.}}{\mathbf{00}}{\mathbf{‐}}{\mathbf{\Omega }}$ resistor alone, what is the percent decrease in current? (d) If the current is kept the same through the combination as it was through the ${\mathbf{10}}{\mathbf{.}}{\mathbf{00}}{\mathbf{‐}}{\mathbf{\Omega }}$ resistor alone, what is the percent increase in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss.

(a) The circuit diagram for the series connection of the ammeter in a resistor circuit is,

(b) The total resistance of the combination is ${\mathrm{R}}_{\mathrm{tot}}=10.02\mathrm{\Omega }$ .

(c) Percentage decrease in the current for the same voltage is role="math" localid="1656396884713" $0.1996%$.

(d) If the current of the combinate is the same the increase in the voltage is $0.2%$

(e) Since the results of current and voltage are very small, they are insignificant.

See the step by step solution

## Step 1: Concept Introduction

The rate of electron flow can be described as the aggregate flow of electrons via a wire. The term "resistance" refers to anything that stands in the way of current flow. To convert electrical energy to light, heat, or movement, an electrical circuit must have resistance.

## Step 2: Circuit Diagram

(a)

The circuit diagram of the connection of an ammeter when a $0.02\mathrm{\Omega }$ ammeter is placed in series with a $10\mathrm{\Omega }$ resistor in a circuit is given below,

Therefore, the circuit diagram is obtained.

## Step 3: Resistance of the combination

(b)

A $0.02\mathrm{\Omega }$ ammeter is placed in series with a $10\mathrm{\Omega }$ resistor in a circuit.

The formula for total resistance is,

${\mathrm{R}}_{\mathrm{combination}}=\mathrm{R}+\mathrm{r}\phantom{\rule{0ex}{0ex}}=10\mathrm{\Omega }+0.02\mathrm{\Omega }\phantom{\rule{0ex}{0ex}}=10.02\mathrm{\Omega }$

Therefore, the value for total resistance is ${\mathrm{R}}_{\mathrm{combination}}=10.02\mathrm{\Omega }$ .

## Step 4: Decrease in the current

(c)

Current through the only resistor is,

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{V}}{10\mathrm{\Omega }}$

And current through combination is,

${\mathrm{I}}_{\mathrm{combination}}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{combination}}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{V}}{10.02\mathrm{\Omega }}$

The decrease in current is,

Therefore, the decrease in current is $0.1996%$.

## Step 5: Increase in voltage

(d)

Voltage through the only resistor is,

$\mathrm{V}=\mathrm{IR}\phantom{\rule{0ex}{0ex}}=\mathrm{I}×10\mathrm{\Omega }$

And voltage through combination is,

${\mathrm{V}}_{\mathrm{Combination}}={\mathrm{IR}}_{\mathrm{Combination}}\phantom{\rule{0ex}{0ex}}=\mathrm{I}×10.02\mathrm{\Omega }$

The increase in voltage is,

${\mathrm{P}}_{\mathrm{I}}=\left(\frac{{\mathrm{V}}_{\mathrm{Combination}}-\mathrm{V}}{\mathrm{V}}\right)×100%\phantom{\rule{0ex}{0ex}}=\left(\frac{\left(\mathrm{I}×10.02\mathrm{\Omega }\right)-\left(\mathrm{I}×10\mathrm{\Omega }\right)}{\left(\mathrm{I}×10\mathrm{\Omega }\right)}\right)×100%\phantom{\rule{0ex}{0ex}}=0.2%$

Therefore, the increase in voltage is $0.2%$ .

## Step 6: Are the changes significant

(e)

The result for the decrease in current is $0.1996%$ .

The result ofthe increase in voltage is $0.2%$ .

These are very small amounts, so they do not affect the circuit in anyway.

Therefore, the results are not significant.