[amp_mcq option1=”50 pF” option2=”100 pF” option3=”150 pF” option4=”200 pF” correct=”option2″]<\/p>\n
The correct answer is B. 100 pF.<\/p>\n
To calculate the equivalent capacitance of two capacitors connected in parallel, we use the following formula:<\/p>\n
$C_{eq} = C_1 + C_2$<\/p>\n
In this case, we are given that $C_1 = 50\\text{ pF}$ and we want to find $C_2$ such that $C_{eq} = 150\\text{ pF}$. Substituting these values into the formula, we get:<\/p>\n
$150\\text{ pF} = 50\\text{ pF} + C_2$<\/p>\n
Solving for $C_2$, we get:<\/p>\n
$C_2 = 150\\text{ pF} – 50\\text{ pF} = 100\\text{ pF}$<\/p>\n
Therefore, the value of capacitance that must be connected in parallel with 50 pF condenser to make an equivalent capacitance of 150 pF is 100 pF.<\/p>\n
The other options are incorrect because they do not result in an equivalent capacitance of 150 pF.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”50 pF” option2=”100 pF” option3=”150 pF” option4=”200 pF” correct=”option2″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[971],"tags":[],"class_list":["post-57109","post","type-post","status-publish","format-standard","hentry","category-electrostatics","no-featured-image-padding"],"yoast_head":"\n