Ammeter and Voltmeter question?

  • Thread starter Thread starter vkash
  • Start date Start date
  • Tags Tags
    Ammeter Voltmeter
AI Thread Summary
When an ammeter and voltmeter are connected in series to a cell, adding a resistance in parallel with the voltmeter affects their readings. The initial assumption that both readings would decrease was challenged, indicating a misunderstanding of the circuit configuration. The correct interpretation involves analyzing how the resistances interact when the new resistance is added. Ultimately, the conclusion reached was that the initial reasoning was incorrect, and the problem was resolved with clarification on the circuit diagram. Understanding the connections is crucial for accurate readings in such circuits.
vkash
Messages
316
Reaction score
1

Homework Statement



A ammeter and a voltmeter are joined in series to a cell . Their reading are A and V respectively. If a resistance is now joined in parallel with voltmeter,
(a) Both A and V will increase
(b) Both A and V will decrease
(c) A will decrease V will increase
(d) A will increase V will decrease

Homework Equations



You know that so i needn't to provide

The Attempt at a Solution



For me option 'B' is correct
why?
because; in first case net resistance in Rv+Ra so current in the circuit is \frac{V}{R<sub>v</sub>+R<sub>a</sub>}[\tex] In second case a new resistance is introduced in the circuit so net resistance is R<sub>v</sub>+R<sub>a</sub>+<b>R</b>. Current in this case is \frac{V}{R&lt;sub&gt;v&lt;/sub&gt;+R&lt;sub&gt;a&lt;/sub&gt;+&lt;b&gt;R&lt;/b&gt;}[\itex]. A we can see quit clearly that \frac{V}{R&amp;lt;sub&amp;gt;v&amp;lt;/sub&amp;gt;+R&amp;lt;sub&amp;gt;a[/sub}[\itex] &amp;amp;gt; \frac{V}{R&amp;amp;lt;sub&amp;amp;gt;v&amp;amp;lt;/sub&amp;amp;gt;+R&amp;amp;lt;sub&amp;amp;gt;a&amp;amp;lt;/sub&amp;amp;gt;+&amp;amp;lt;b&amp;amp;gt;R&amp;amp;lt;/b&amp;amp;gt;}&amp;lt;br /&amp;gt; so current in second case (say I&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;) is greater than current in first case(say I&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;) &amp;lt;br /&amp;gt; Potential difference between the voltmeter in first case is R&amp;amp;lt;sub&amp;amp;gt;v&amp;amp;lt;/sub&amp;amp;gt;I&amp;amp;lt;sub&amp;amp;gt;1&amp;amp;lt;/sub&amp;amp;gt;[\itex] and in second case is R&amp;amp;amp;lt;sub&amp;amp;amp;gt;v&amp;amp;amp;lt;/sub&amp;amp;amp;gt;I&amp;amp;amp;lt;sub&amp;amp;amp;gt;2&amp;amp;amp;lt;/sub&amp;amp;amp;gt;[\itex]. I&amp;amp;amp;lt;sub&amp;amp;amp;gt;2&amp;amp;amp;lt;/sub&amp;amp;amp;gt;&amp;amp;amp;amp;lt;i&amp;amp;amp;lt;sub&amp;amp;amp;gt;1&amp;amp;amp;lt;/sub&amp;amp;amp;gt; so potential difference between the voltmeter in first case should greater than that in second case.&amp;amp;amp;lt;br /&amp;amp;amp;gt; so what&amp;amp;amp;amp;#039;s the conclusion; that is ammeter reading as well as voltmeter reading will drop so. Option B is correct for me.&amp;amp;amp;lt;br /&amp;amp;amp;gt; As always i ask. &amp;amp;amp;lt;b&amp;amp;amp;gt; where am i wrong?&amp;amp;amp;lt;/b&amp;amp;amp;gt;&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;b&amp;amp;amp;gt;sorry friends i tried to change / \ and some others, but none work with tex itex so try understand as it is written or just put answer.&amp;amp;amp;lt;/b&amp;amp;amp;gt;&amp;lt;/sub&amp;gt;
 
Last edited:
Physics news on Phys.org
vkash said:

Homework Statement



A ammeter and a voltmeter are joined in series to a cell . Their reading are A and V respectively. If a resistance is now joined in parallel with voltmeter,
(a) Both A and V will increase
(b) Both A and V will decrease
(c) A will decrease V will increase
(d) A will increase V will decreaseFor me option 'B' is correct
why?
because; in first case net resistance in Rv+Ra so current in the circuit is \frac{V}{R<sub>v</sub>+R<sub>a</sub>}[\tex] In second case a new resistance is introduced in the circuit <b>so net resistance is R<sub>v</sub>+R<sub>a</sub>+<b>R</b></b>.
<br /> <br /> It is wrong.<br /> <br /> Read the problem again, especially the text in blue. Draw the circuit diagram, and find out how the resistances are connected. <br /> <br /> ehild
 
Last edited:
ehild said:
It is wrong.

Read the problem again, especially the text in blue. Draw the circuit diagram, and find out how the resistances are connected.

ehild

:smile:thanks buddy:smile:. Question solved.:smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top