JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 24)
In the given circuit, an ideal voltmeter
connected across the 10$$\Omega $$ resistance reads 2V.
The internal resistance r, of each cell is:
_10th_April_Morning_Slot_en_24_1.png)
1.5 $$\Omega $$
1 $$\Omega $$
0.5 $$\Omega $$
0 $$\Omega $$
Explanation
For the given circuit
_10th_April_Morning_Slot_en_24_2.png)
Given that VAB = 2 V
$$ \therefore $$ I = $${2 \over {15}} + {2 \over {10}} = {1 \over 3}A$$
Also I(2r + 2) = 1.5 + 1.5 – VAB
$$ \Rightarrow $$ 2r + 2 = (3–2)3
$$ \Rightarrow $$ $$r = {1 \over 2}\Omega $$
Given that VAB = 2 V$$ \therefore $$ I = $${2 \over {15}} + {2 \over {10}} = {1 \over 3}A$$
Also I(2r + 2) = 1.5 + 1.5 – VAB
$$ \Rightarrow $$ 2r + 2 = (3–2)3
$$ \Rightarrow $$ $$r = {1 \over 2}\Omega $$
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