During Ohm’s Law experiment we hold the voltmeter across the resistor and ammeter in series with the circuit. Explain why.
Added by Angela F.
Step 1
This is important because Ohm's Law states that the voltage across a resistor is directly proportional to the current flowing through it. ** Show more…
Show all steps
Your feedback will help us improve your experience
Ankur S and 84 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Why do we have to connect a voltmeter in parallel and an ammeter in series
Pritesh R.
D3. Voltmeter Usage Explain why a voltmeter is connected in parallel with the resistor. In order for the voltmeter to have little impact on the circuit, what should its resistance be? Explain. {Hint: I am not looking for a number answer. Think about what the current does when another branch is added to a node – how do you minimize that?} D4. Ammeter Usage Explain why an ammeter is connected in series with the circuit. In order for an ammeter to have little impact on the circuit, what should its resistance be? Explain. {Hint: I am not looking for a number answer. Think about the voltage drop created by current through a component. What happens when you add another component to a loop – how do you minimize that?}
Shaiju T.
A voltmeter (of resistance $\left.R_{\Delta \mathrm{V}}\right)$ and an ammeter (of resistance $\left.R_{\mathrm{A}}\right)$ are connected to measure a resistance $R$, as in Fig. $27-30 a$. The resistance is given by $R=\Delta V / i$, where $\Delta V$ is the voltmeter reading and $i$ is the current in the resistance $R$. Some of the current $i^{\prime}$ registered by the ammeter goes through the voltmeter, so that the ratio of the meter readings $\left(=\Delta V / i^{\prime}\right)$ gives only an apparent resistance reading $R^{\prime}$. Show that $R$ and $R^{\prime}$ are related by $$ \frac{1}{R}=\frac{1}{R^{\prime}}-\frac{1}{R_{\Delta \mathrm{V}}} $$ Note that as $R_{\Delta \mathrm{V}} \rightarrow \infty, R^{\prime} \rightarrow R$. Ignore $R_{0}$ for now.
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD