An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu
1 answer:
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The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
where q is the charge of the proton,
, with 
being the elementary charge, and 
and 
are the initial and final voltage.
Substituting, we get (in electronvolts):
and in Joule:
where q is the charge of the proton,
Substituting, we get (in electronvolts):
and in Joule:
Answer:
v = -6m/s
Explanation:
For t = 1:
Answer:
5 ohms
Explanation:
Given:
EMF of the ideal battery (E) = 60 V
Voltage across the terminals of the battery (V) = 40 V
Current across the terminals (I) = 4 A
Let the internal resistance be 'r'.
Now, we know that, the voltage drop in the battery is given as:
Therefore, the voltage across the terminals of the battery is given as:
Now, rewriting in terms of 'r', we get:
Plug in the given values and solve for 'r'. This gives,
Therefore, the internal resistance of the battery is 5 ohms.